question_id stringlengths 8 35 | subject stringclasses 3
values | chapter stringclasses 90
values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
values | explanation stringlengths 1 10.6k | question_type stringclasses 3
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jaoe38c1lsflxsi0 | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>The bob of a pendulum was released from a horizontal position. The length of the pendulum is $$10 \mathrm{~m}$$. If it dissipates $$10 \%$$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:</p>
<p>[Use, $$\mathrm{g}: 10 \mathrm{~ms}^{-2}$$]</p> | [{"identifier": "A", "content": "$$5 \\sqrt{6} \\mathrm{~ms}^{-1}$$\n"}, {"identifier": "B", "content": "$$5 \\sqrt{5} \\mathrm{~ms}^{-1}$$\n"}, {"identifier": "C", "content": "$$2 \\sqrt{5} \\mathrm{~ms}^{-1}$$\n"}, {"identifier": "D", "content": "$$6 \\sqrt{5} \\mathrm{~ms}^{-1}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr7le3s/5e7d48b1-ff39-4026-8060-31f894d71fa9/88e4b280-ce32-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr7le3t.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr7le3s/5e7d48b1-ff39-4026-8060-31f894d71fa9/88e4b280-ce32-11ee... | mcq | jee-main-2024-online-29th-january-evening-shift | 12,790 |
vFMPNzyqwdwRNs7q | physics | units-and-measurements | dimensions-of-physical-quantities | Identify the pair whose dimensions are equal | [{"identifier": "A", "content": "torque and work"}, {"identifier": "B", "content": "stress and energy"}, {"identifier": "C", "content": "force and stress"}, {"identifier": "D", "content": "force and work"}] | ["A"] | null | <p>To determine which pairs have equivalent dimensions, we examine the respective dimensional formulas.</p>
<p>For work (W) specified as $ \overrightarrow{F} \cdot \overrightarrow{s} $:</p>
<p>$ W = F s \cos \theta $</p>
<p>The dimensions are:</p>
<p>$ W = [MLT^{-2}][L] = [ML^2T^{-2}] $</p>
<p>For torque ($\overri... | mcq | aieee-2002 | 12,791 |
07Fb9a0w1SWCOyMV | physics | units-and-measurements | dimensions-of-physical-quantities | Dimensions of $${1 \over {{\mu _0}{\varepsilon _0}}}$$, where symbols have their usual meaning, are | [{"identifier": "A", "content": "[ L<sup>-1</sup>T ]"}, {"identifier": "B", "content": "[ L<sup>-2</sup>T<sup>2</sup> ]"}, {"identifier": "C", "content": "[ L<sup>2</sup>T<sup>-2</sup> ]"}, {"identifier": "D", "content": "[ LT<sup>-1</sup> ]"}] | ["C"] | null | The velocity of light in vacuum is <br><br>
c = $${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$; <br><br>
$$\therefore[{1 \over {{\mu _0}{\varepsilon _0}}}]$$ = [c<sup>2</sup>] = [L<sup>2</sup>T<sup>-2</sup>]
<br><br>$$\therefore$$ Dimension of $${1 \over {{\mu _0}{\varepsilon _0}}}$$ = [L<sup>2</sup>T<sup>-2</sup>] | mcq | aieee-2003 | 12,792 |
02536sILFKVDSkVh | physics | units-and-measurements | dimensions-of-physical-quantities | The physical quantities not having same dimensions are | [{"identifier": "A", "content": "torque and work"}, {"identifier": "B", "content": "momentum and Planck's constant"}, {"identifier": "C", "content": "stress and Young's modulus"}, {"identifier": "D", "content": "speed and $${\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}}$$"}] | ["B"] | null | Momentum = mv = [$${M{L}{T^{ - 1}}}$$]<br><br>
Planck's constant, h = $${E \over v}$$ = $${[{M{L^2}{T^{ - 2}]}} \over {[{T^{ - 1}}]}}$$ = $$[{M{L^2}{T^{ - 1}}}]$$
<br><br>So Momentum and Planck's constant do not have same dimensions. | mcq | aieee-2003 | 12,793 |
vKN4F98zXejlgqUZ | physics | units-and-measurements | dimensions-of-physical-quantities | The dimension of magnetic field in M, L, T and C (coulomb) is given as | [{"identifier": "A", "content": "MLT<sup>-1</sup>C<sup>-1</sup>"}, {"identifier": "B", "content": "MT<sup>2</sup>C<sup>-2</sup>"}, {"identifier": "C", "content": "MT<sup>-1</sup>C<sup>-1</sup>"}, {"identifier": "D", "content": "MT<sup>-2</sup>C<sup>-1</sup>"}] | ["C"] | null | We know that, <br>Lorentz force $$\left| {\overrightarrow F } \right| = \left| {q\overrightarrow v \times \overrightarrow B } \right|$$<br><br>
$$\therefore [B] $$ = $${[F] \over {[q][v]}}$$ = $${[{ML{T^{ - 2}}]} \over {[C] \times [L{T^{ - 1}}]}}$$ = [$$M{T^{ - 1}}{C^{ - 1}}$$] | mcq | aieee-2008 | 12,796 |
JsYfxzWCG2nBVfIp | physics | units-and-measurements | dimensions-of-physical-quantities | Let [$${\varepsilon _0}$$] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time
and A = electric current, then: | [{"identifier": "A", "content": "$${\\varepsilon _0} = \\left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \\right]$$"}, {"identifier": "B", "content": "$${\\varepsilon _0} = $$$$\\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \\right]$$"}, {"identifier": "C", "content": "$${\\varepsilon _0} = \\left[ {{M^1}{L^2}{T^1}{A^2}} \\right]$$"}, {"... | ["B"] | null | From Coulomb's law we know,
<br><br>$$F = {1 \over {4\pi { \in _0}}}{{{q_1}{q_2}} \over {{r^2}}}$$
<br><br>$$\therefore$$ $${ \in _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {F{r^2}}}$$
<br><br>Hence, $$\left[ {{ \in _0}} \right] = {{\left[ {AT} \right]\left[ {AT} \right]} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^... | mcq | jee-main-2013-offline | 12,797 |
a0Oc15SkQSAm3RCpyzXhr | physics | units-and-measurements | dimensions-of-physical-quantities | In the following ‘I’ refers to current and other symbols have their usual meaning.
Choose the option that corresponds to the <sup></sup>dimensions of electrical conductivity : | [{"identifier": "A", "content": "ML<sup>$$-$$3</sup> T<sup>$$-$$3</sup> I2"}, {"identifier": "B", "content": "M<sup>$$-$$1</sup> L<sup>3</sup> T<sup>3</sup> I"}, {"identifier": "C", "content": "M<sup>$$-$$1</sup> L<sup>$$-$$3</sup> T<sup>3</sup> I<sup>2</sup> "}, {"identifier": "D", "content": "M<sup>$$-$$1</sup> L<su... | ["C"] | null | We know. resistivity ($$\rho $$) = $${{RA} \over L}$$
<br><br>and conductivity = $${1 \over \rho }$$ = $${1 \over {RA}}$$
<br><br>As R = $${V \over {\rm I}}$$
<br><br>$$ \therefore $$ conductivity = $${{L{\rm I}} \over {VA}}$$
<br><br>Also V = $${\omega \over q}$$ = $${\ome... | mcq | jee-main-2016-online-9th-april-morning-slot | 12,798 |
JKiVgY5u6sdsLyeSDGYdB | physics | units-and-measurements | dimensions-of-physical-quantities | A, B, C and D are four different physical quantities having different
dimensions. None of them is dimensionless. But we know that the equation AD = C ln (BD) holds true. Then which of the combination is <b>not</b> a meaningful quantity ? | [{"identifier": "A", "content": "A<sup>2</sup> $$-$$ B<sup>2</sup>C<sup>2</sup>"}, {"identifier": "B", "content": "$${{\\left( {A - C} \\right)} \\over D}$$"}, {"identifier": "C", "content": "$${A \\over B} - C$$ "}, {"identifier": "D", "content": "$${C \\over {BD}} - {{A{D^2}} \\over C}$$"}] | null | null | Given,
<br><br>AD = c ln (BD)
<br><br>As log is dimensionless.
<br><br>So, [BD] = 1 $$ \Rightarrow $$ [B] = $${1 \over {\left[ D \right]}}$$
<br><br>And [AD] = [C]
<br><br>Now checking options one by one
<br><br>(a) [B<sup>2</sup> C<sup>2</sup>] = [B<sup>2</sup>] [A... | mcqm | jee-main-2016-online-10th-april-morning-slot | 12,799 |
2lYPc9yvA2f6HHSJvpQTT | physics | units-and-measurements | dimensions-of-physical-quantities | The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. <br/><br/>Which of the following correctly gives the Planck length ? | [{"identifier": "A", "content": "G $$\\hbar $$<sup>2</sup> c<sup>3</sup>"}, {"identifier": "B", "content": "G<sup>2</sup> $$\\hbar $$ c"}, {"identifier": "C", "content": "$${G^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}{\\hbar ^2}c$$"}, {"identifier": "... | ["D"] | null | Plank length,
<br><br>$$\ell $$ = k G<sup>p</sup> $$\hbar $$<sup>q</sup> C<sup>r</sup>
<br><br>[ M<sup>o</sup> L T<sup>o</sup>] = [ M<sup>$$-$$1</sup> L<sup>3</sup> T<sup>$$-$$2</sup> ]<sup>p</sup> [ M L<sup>2</sup> T<sup>$$-$$1</sup>] <sup>q</sup> [ L T<sup>$$-$$1</sup>]<sup>r</sup>
<br><br>[M<sup>o</sup> L T<sup>o<... | mcq | jee-main-2018-online-15th-april-evening-slot | 12,801 |
YjHRHYhwFh2vTu0gJtLrN | physics | units-and-measurements | dimensions-of-physical-quantities | If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be: | [{"identifier": "A", "content": "V<sup>$$-$$2</sup>A<sup>2</sup>F<sup>2</sup>"}, {"identifier": "B", "content": "V<sup>$$-$$4</sup>A<sup>$$-$$2</sup>F"}, {"identifier": "C", "content": "V<sup>$$-$$4</sup>A<sup>2</sup>F"}, {"identifier": "D", "content": "V<sup>$$-$$2</sup>A<sup>2</sup>F<sup>$$-$$2</sup>"}] | ["C"] | null | We know,
<br><br>Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$
<br><br>$$ \therefore $$ [Y] = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$$ = [ ML<sup>-1</sup>T<sup>-2</sup>]
<br><br>Let [Y] = [V]<sup>x</sup> [A]<sup>y</sup> [F]<sup>z</sup>
<br><br>$$ \therefore $$ [ ML<s... | mcq | jee-main-2019-online-11th-january-evening-slot | 12,802 |
TSlvOuUar7yzKEyuix3rsa0w2w9jx3p37sc | physics | units-and-measurements | dimensions-of-physical-quantities | Which of the following combinations has the dimension of electrical resistance ($$ \in $$<sub>0 </sub>is the permittivity of
vacuum and $$\mu $$<sub>0</sub> is the permeability of vacuum)? | [{"identifier": "A", "content": "$$\\sqrt {{{{ \\in _0}} \\over {{\\mu _0}}}} $$"}, {"identifier": "B", "content": "$${{{{ \\in _0}} \\over {{\\mu _0}}}}$$"}, {"identifier": "C", "content": "$$\\sqrt {{{{\\mu _0}} \\over {{ \\in _0}}}} $$"}, {"identifier": "D", "content": "$${{{{\\mu _0}} \\over {{ \\in _0}}}}$$"}] | ["C"] | null | According to Coulomb's law
<br><br>F = $${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$$
<br><br>$$ \therefore $$ $${ \in _0} = {1 \over {4\pi }}{{{q^2}} \over {F{r^2}}}$$
<br><br>$$\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}$$ = $$\left[ {{... | mcq | jee-main-2019-online-12th-april-morning-slot | 12,803 |
IkxrAnrmblHnX3kEXK3rsa0w2w9jwz162jd | physics | units-and-measurements | dimensions-of-physical-quantities | In the formula X = 5YZ<sup>2</sup>
, X and Z have dimensions of capacitance and magnetic field, respectively. What are
the dimensions of Y in SI units? | [{"identifier": "A", "content": "[M<sup>\u20133</sup>L<sup>\u20132</sup>T<sup>8</sup>A<sup>4</sup>]"}, {"identifier": "B", "content": "[M<sup>\u20132</sup>L<sup>\u20132</sup>T<sup>6</sup>A<sup>3</sup>]"}, {"identifier": "C", "content": "[M<sup>\u20131</sup>L<sup>\u20132</sup>T<sup>4</sup>A<sup>2</sup>]"}, {"identifier"... | ["A"] | null | Capacitance (C) = $${{{Q^2}} \over {2E}}$$ = $${{\left[ {{A^2}{T^2}} \right]} \over {\left[ {M{L^2}{T^{ - 2}}} \right]}}$$ = $$\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]$$ X
<br><br>Magnetic field (B) = $${F \over {IL}}$$ = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ A \right]\left[ L \right]}}$$ = $${\left[ ... | mcq | jee-main-2019-online-10th-april-evening-slot | 12,804 |
0DIhSkOruKebYWI5WN7Ko | physics | units-and-measurements | dimensions-of-physical-quantities | If surface tension (S), Moment of inertia (I) and
Planck's constant (h), were to be taken as the
fundamental units, the dimensional formula for
linear momentum would be :- | [{"identifier": "A", "content": "S<sup>1/2</sup>I<sup>1/2</sup>h<sup>0</sup>"}, {"identifier": "B", "content": "S<sup>3/2</sup>I<sup>1/2</sup>h<sup>0</sup>"}, {"identifier": "C", "content": "S<sup>1/2</sup>I<sup>1/2</sup>h<sup>-1</sup>"}, {"identifier": "D", "content": "S<sup>1/2</sup>I<sup>3/2</sup>h<sup>-1</sup>"}] | ["A"] | null | We know,
<br><br>surface tension (S) = $${F \over L}$$ = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ L \right]}}$$
<br><br>$$ \therefore $$ [S] = $${\left[ {M{T^{ - 2}}} \right]}$$
<br><br>Moment of inertia (I) = mr<sup>2</sup>
<br><br>$$ \therefore $$ [I] = $${\left[ {M{L^2}} \right]}$$
<br><br>Planck's constant ... | mcq | jee-main-2019-online-8th-april-evening-slot | 12,805 |
At9P01Is6s69OyEoECX4d | physics | units-and-measurements | dimensions-of-physical-quantities | The force of interaction between two atoms is given by F = $$\alpha $$$$\beta $$exp $$\left( { - {{{x^2}} \over {\alpha kt}}} \right)$$; where x is the distance, k is the Boltzmann constant and T is temperature and $$\alpha $$ and $$\beta $$ are two constants. The dimension of $$\beta $$ is : | [{"identifier": "A", "content": "M<sup>2</sup>L<sup>2</sup>T<sup>$$-$$2</sup>"}, {"identifier": "B", "content": "M<sup>2</sup>LT<sup>$$-$$4</sup>"}, {"identifier": "C", "content": "MLT<sup>$$-$$4</sup>"}, {"identifier": "D", "content": "M<sup>0</sup>L<sup>2</sup>LT<sup>$$-$$4</sup>"}] | ["B"] | null | $$F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}$$
<br><br>$$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$$
<br><br>$${{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}$$ $$=$$ $${M^o}{L^o}{T^o}$$
<br><br>$$ \Rightarrow $$ $$\left[ \alpha \right] = {M^{ - 1}}{... | mcq | jee-main-2019-online-11th-january-morning-slot | 12,808 |
f9L3qCp6iK7QkgvEWnmj6 | physics | units-and-measurements | dimensions-of-physical-quantities | Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to : | [{"identifier": "A", "content": "$$\\sqrt {{{h{c^5}} \\over G}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{{{c^3}} \\over {Gh}}} $$"}, {"identifier": "C", "content": "$$\\sqrt {{{Gh} \\over {{c^5}}}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{{Gh} \\over {{c^3}}}} $$"}] | ["C"] | null | Let t $$ \propto $$ G<sup>x</sup> h<sup>y</sup> c<sup>z</sup>
<br><br>$$ \therefore $$ [t] = [G]<sup>x</sup> [h]<sup>y</sup> [c]<sup>z</sup> . . . . . (1)
<br><br>We know,
<br><br>F = $${{G{M^2}} \over {{R^2}}}$$
<br><br>$$ \Rightarrow $$ G = $${{F{R^2}} \over {{M^2}}}$$... | mcq | jee-main-2019-online-9th-january-evening-slot | 12,809 |
VvLRXRNWzR6zCBsBJt7k9k2k5gvtufp | physics | units-and-measurements | dimensions-of-physical-quantities | The dimension of stopping potential V<sub>0</sub> in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is : | [{"identifier": "A", "content": "h<sup>1/3</sup> G<sup>2/3</sup> c<sup>1/3</sup> A<sup>\u20131</sup>"}, {"identifier": "B", "content": "h<sup>0</sup> c<sup>5</sup> G<sup>-1</sup> A<sup>-1</sup>"}, {"identifier": "C", "content": "h<sup>2/3</sup> c<sup>5/3</sup> G<sup>1/3</sup> A<sup>\u20131</sup>"}, {"identifier": "D", ... | ["B"] | null | V<sub>0</sub> $$ \propto $$ h<sup>P</sup>c<sup>Q</sup>G<sup>R</sup>I<sup>S</sup>
<br><br>[V<sub>0</sub>] = [M<sup>1</sup>L<sup>2</sup>T<sup>–3</sup>A<sup>–1</sup>]
<br><br>[c] = [L<sup>1</sup>T<sup>–1</sup>]
<br><br>[h] = [M<sup>1</sup>L<sup>2</sup>T<sup>–1</sup>]
<br><br>[G] = [M<sup>–1</sup>L<sup>3</sup>T<sup>–2</sup... | mcq | jee-main-2020-online-8th-january-morning-slot | 12,810 |
GziM0gxsRQMxENruUgjgy2xukfl1yz9k | physics | units-and-measurements | dimensions-of-physical-quantities | The quantities x = $${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$, y = $${E \over B}$$ and z = $${l \over {CR}}$$ are
<br/>defined where C-capacitance, R-Resistance,
l-length, E-Electric field, B-magnetic field and
$${{\varepsilon _0}}$$, $${{\mu _0}}$$, - free space permittivity and permeability
respectively. Then ... | [{"identifier": "A", "content": "Only y and z have the same dimension"}, {"identifier": "B", "content": "x, y and z have the same dimension"}, {"identifier": "C", "content": "Only x and y have the same dimension"}, {"identifier": "D", "content": "Only x and z have the same dimension"}] | ["B"] | null | x = $${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$ = speed
<br><br>$$ \therefore $$ [x] = [L<sup>1</sup>T<sup>–1</sup>]
<br><br>y = $${E \over B}$$ = speed
<br><br>$$ \therefore $$ [y] = [L<sup>1</sup>T<sup>–1</sup>]
<br><br>z = $${l \over {CR}}$$ = $${l \over \tau }$$
<br><br>$$ \Rightarrow $$ [z] = [L<sup>1</sup>T... | mcq | jee-main-2020-online-5th-september-evening-slot | 12,811 |
7xGT51VK2ieW0N0ftKjgy2xukfamk105 | physics | units-and-measurements | dimensions-of-physical-quantities | A quantity x is given by $$\left( {{{IF{v^2}} \over {W{L^4}}}} \right)$$ in terms of moment of inertia I, force F, velocity v, work W and
Length L. The dimensional formula for x is same as that of :
| [{"identifier": "A", "content": "Coefficient of viscosity"}, {"identifier": "B", "content": "Force constant"}, {"identifier": "C", "content": "Energy density"}, {"identifier": "D", "content": "Planck's constant"}] | ["C"] | null | x = $$\left( {{{IF{v^2}} \over {W{L^4}}}} \right)$$
<br><br>$$ \therefore $$ [x] = $${{\left[ {M{L^2}} \right]\left[ {ML{T^{ - 2}}} \right]{{\left[ {L{T^{ - 1}}} \right]}^2}} \over {\left[ {M{L^2}{T^{ - 2}}} \right]{{\left[ L \right]}^4}}}$$
<br><br>= [ML<sup>-1</sup>T<sup>-2</sup>]
<br><br>= [Energy density] | mcq | jee-main-2020-online-4th-september-evening-slot | 12,812 |
tKdPgxYEygdJoj7jfFjgy2xukf6j18rx | physics | units-and-measurements | dimensions-of-physical-quantities | Dimensional formula for thermal conductivity is (here K denotes the temperature): | [{"identifier": "A", "content": "MLT<sup>\u20133</sup>K<sup>\u20131</sup>"}, {"identifier": "B", "content": "MLT<sup>\u20132</sup>K<sup>\u20132</sup>"}, {"identifier": "C", "content": "MLT<sup>\u20132</sup>K"}, {"identifier": "D", "content": "MLT<sup>\u20133</sup>K"}] | ["A"] | null | $$ \therefore $$ $${{d\theta } \over {dt}} = kA{{dT} \over {dx}}$$
<br><br>$$ \Rightarrow $$ k = $${{\left( {{{d\theta } \over {dt}}} \right)} \over {A\left( {{{dT} \over {dx}}} \right)}}$$
<br><br>$$ \Rightarrow $$ [k] = $${{\left[ {M{L^2}{T^{ - 3}}} \right]} \over {\left[ {{L^2}} \right]\left[ {K{L^{ - 1}}} \right]}}... | mcq | jee-main-2020-online-4th-september-morning-slot | 12,813 |
nS9LaEzHUKxeuFS3z1jgy2xukexsdb2t | physics | units-and-measurements | dimensions-of-physical-quantities | If momentum (P), area (A) and time (T) are
taken to be the fundamental quantities then the
dimensional formula for energy is | [{"identifier": "A", "content": "[P<sup>2</sup>AT<sup>\u20132</sup>]"}, {"identifier": "B", "content": "$$\\left[ {{P^{{1 \\over 2}}}A{T^{ - 1}}} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {P{A^{{1 \\over 2}}}{T^{ - 1}}} \\right]$$"}, {"identifier": "D", "content": "[PA<sup>\u20131</sup>T<sup>\u20132</sup>... | ["C"] | null | Let
[E] =
K[P]<sup>x</sup>[A]<sup>y</sup> [T]<sup>z</sup>
<br><br> [ML<sup>2</sup>T<sup>–2</sup>] = [MLT<sup>–1</sup>]<sup>x</sup>[L<sup>2</sup>]<sup>y</sup>[T]<sup>z</sup>
<br><br> [ML<sup>2</sup>T<sup>–2</sup>] = [M<sup>x</sup>][L<sup>x+2y</sup>][T<sup>–x+z</sup>]
<br><br>Comparing both side we get,
<br><br> x = ... | mcq | jee-main-2020-online-2nd-september-evening-slot | 12,815 |
n5mZ89v3oYDnfYyg0Ujgy2xukev18nyc | physics | units-and-measurements | dimensions-of-physical-quantities | If speed V, area A and force F are chosen as
fundamental units, then the dimension of
Young’s modulus will be | [{"identifier": "A", "content": "FA<sup>\u20131</sup>V<sup>0</sup>"}, {"identifier": "B", "content": "FA<sup>2</sup>V<sup>\u20131</sup>"}, {"identifier": "C", "content": "FA<sup>2</sup>V<sup>\u20132</sup>"}, {"identifier": "D", "content": "FA<sup>2</sup>V<sup>\u20133</sup>"}] | ["A"] | null | Y = k [F]<sup>x</sup>
[A]<sup>y</sup>
[V]<sup>z</sup>
<br><br>[M<sup>1</sup>L<sup>1</sup>T
<sup>–2</sup>] = [MLT<sup>–2</sup>]<sup>x</sup> [L<sup>2</sup>]<sup>y</sup> [LT<sup>–1</sup>]<sup>z</sup>
<br><br>[M<sup>1</sup>L<sup>1</sup>T
<sup>–2</sup>] = [M]<sup>x</sup> [L]<sup>x+2y+z</sup>[T]<sup>–2x–z </sup>
<br><br>Co... | mcq | jee-main-2020-online-2nd-september-morning-slot | 12,816 |
qPVBPF82REXPyKsffg7k9k2k5i6hb7m | physics | units-and-measurements | dimensions-of-physical-quantities | A quantity f is given by $$f = \sqrt {{{h{c^5}} \over G}} $$ where c is
speed of light, G universal gravitational
constant and h is the Planck's constant.
Dimension of f is that of : | [{"identifier": "A", "content": "Energy"}, {"identifier": "B", "content": "Momentum"}, {"identifier": "C", "content": "Area"}, {"identifier": "D", "content": "Volume"}] | ["A"] | null | [h] = M<sup>1</sup>L<sup>2</sup>T<sup>–1</sup>
<br>[C] = L<sup>1</sup>T<sup>–1</sup>
<br>[G] = M<sup>–1</sup>L<sup>3</sup>T<sup>–2</sup>
<br><br>[f] = $$\sqrt {{{M{L^2}{T^{ - 1}} \times {L^5}{T^{ - 5}}} \over {{M^{ - 1}}{L^3}{T^{ - 2}}}}} $$ = M<sup>1</sup>L<sup>2</sup>T<sup>–2</sup> | mcq | jee-main-2020-online-9th-january-morning-slot | 12,817 |
xEjU6Vvvmxsi4oBcxp7k9k2k5farxr8 | physics | units-and-measurements | dimensions-of-physical-quantities | The dimension of $${{{B^2}} \over {2{\mu _0}}}$$, where B is magnetic field and $${{\mu _0}}$$
is the magnetic permeability of vacuum,
is : | [{"identifier": "A", "content": "ML<sup>2</sup>T<sup>\u20132</sup>"}, {"identifier": "B", "content": "MLT<sup>\u20132</sup>"}, {"identifier": "C", "content": "ML<sup>-1</sup>T<sup>\u20132</sup>"}, {"identifier": "D", "content": "ML<sup>2</sup>T<sup>\u20131</sup>"}] | ["C"] | null | As $${{{B^2}} \over {2{\mu _0}}}$$ = Energy per unit volume
<br><br>$$ \therefore $$ Dimension = $${{M{L^2}{T^{ - 2}}} \over {{L^3}}}$$ = ML<sup>-1</sup>T<sup>–2</sup> | mcq | jee-main-2020-online-7th-january-evening-slot | 12,818 |
wkrPtWXp8dheOeww7O1klri6brz | physics | units-and-measurements | dimensions-of-physical-quantities | The work done by a gas molecule in an isolated system is <br/><br/>given by, $$W = \alpha {\beta ^2}{e^{ - {{{x^2}} \over {\alpha kT}}}}$$, where x is the displacement, k is the Boltzmann constant and T is the temperature. $$\alpha$$ and $$\beta$$ are constants. Then the dimensions of $$\beta$$ will be : | [{"identifier": "A", "content": "$$[{M^0}L{T^0}]$$"}, {"identifier": "B", "content": "$$[M{L^2}{T^{ - 2}}]$$"}, {"identifier": "C", "content": "$$[ML{T^{ - 2}}]$$"}, {"identifier": "D", "content": "$$[{M^2}L{T^2}]$$"}] | ["C"] | null | where, k is Boltzmann constant,<br/><br/>T is temperature and x is displacement.<br/><br/>We know that, $${{{x^2}} \over {\alpha kT}}$$ is a dimensionless quantity.<br/><br/>$$\therefore$$ $$\left[ {{{{x^2}} \over {\alpha kT}}} \right] = [{M^0}{L^0}{T^0}] \Rightarrow [\alpha ] = {{[{x^2}]} \over {[k][T]}}$$<br/><br/>$$... | mcq | jee-main-2021-online-24th-february-morning-slot | 12,819 |
uInD6uieRdDlLVE1Vj1klrx8pev | physics | units-and-measurements | dimensions-of-physical-quantities | Match List - I with List - II :<br/><br/><table>
<thead>
<tr>
<th></th>
<th>List I</th>
<th></th>
<th>List II</th>
</tr>
</thead>
<tbody>
<tr>
<td>(a)</td>
<td>h (Planck's constant)</td>
<td>(i)</td>
<td>$$[ML{T^{ - 1}}]$$</td>
</tr>
<tr>
<td>(b)</td>
<td>E (kinetic energy)</td>
<td>(ii)</td>
<td>$$[M{L^2}{T^{ - 1}}]$$... | [{"identifier": "A", "content": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (i)"}, {"identifier": "B", "content": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)"}, {"identifier": "C", "content": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (ii), (c) $$ \... | ["A"] | null | Kinetic Energy,<br><br>$${1 \over 2}m{v^2} = [M{L^2}{T^{ - 2}}]$$<br><br>Momentum,<br><br>$$mv = [ML{T^{ - 1}}]$$<br><br>Plank constant :<br><br>$$E = h\gamma $$<br><br>$$ \Rightarrow M{L^2}{T^{ - 2}} = h \times {1 \over T}$$<br><br>$$ \Rightarrow h = [M{L^2}{T^{ - 1}}]$$<br><br>Also, $$E = qV$$<br><br>$$ \Rightarrow V... | mcq | jee-main-2021-online-25th-february-morning-slot | 12,820 |
NTDwUXVnzYf9Vp17zd1klt344f5 | physics | units-and-measurements | dimensions-of-physical-quantities | If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity $${1 \over {4\pi {\varepsilon _0}}}{{|e{|^2}} \over {hc}}$$ has dimensions of : | [{"identifier": "A", "content": "$$[ML{T^{ - 1}}]$$"}, {"identifier": "B", "content": "$$[ML{T^0}]$$"}, {"identifier": "C", "content": "$$[{M^0}{L^0}{T^0}]$$"}, {"identifier": "D", "content": "$$[L{C^{ - 1}}]$$"}] | ["C"] | null | Given <br><br>e = electronic charge<br><br>c = speed of light in free space<br><br>h = Planck's constant
<br><br>We know, E = $${{hc} \over \lambda }$$
<br><br>and $$F = {1 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {{d^2}}}$$ $$ \Rightarrow $$ $${{{q^2}} \over {4\pi {\varepsilon _0}}} = F{d^2}$$
<br><br>$${1 \over {... | mcq | jee-main-2021-online-25th-february-evening-slot | 12,821 |
Wgp580uhw4Gc7CC3Si1kltil3pw | physics | units-and-measurements | dimensions-of-physical-quantities | In a typical combustion engine the workdone by a gas molecule is given by $$W = {\alpha ^2}\beta {e^{{{ - \beta {x^2}} \over {kT}}}}$$, where x is the displacement, k is the Boltzmann constant and T is the temperature. If $$\alpha$$ and $$\beta$$ are constants, dimensions of $$\alpha$$ will be : | [{"identifier": "A", "content": "$$[{M^0}L{T^0}]$$"}, {"identifier": "B", "content": "$$[ML{T^{ - 1}}]$$"}, {"identifier": "C", "content": "$$[ML{T^{ - 2}}]$$"}, {"identifier": "D", "content": "$$[{M^2}L{T^{ - 2}}]$$"}] | ["A"] | null | kT has dimension of energy<br><br>$${{\beta {x^2}} \over {kT}}$$ is dimensionless<br><br>$$[\beta ][{L^2}] = [M{L^2}{T^{ - 2}}]$$<br><br>$$[\beta ] = [M{T^{ - 2}}]$$<br><br>$${\alpha ^2}\beta $$ has dimensions of work<br><br>$$[{\alpha ^2}][M{T^{ - 2}}] = [M{L^2}{T^{ - 2}}]$$<br><br>$$[\alpha ] = [{M^0}L{T^0}]$$ | mcq | jee-main-2021-online-26th-february-morning-slot | 12,822 |
11GkXBqmyuMqTqOwvf1kluld119 | physics | units-and-measurements | dimensions-of-physical-quantities | If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of $$\lambda$$ where C/V = $$\lambda$$ ? | [{"identifier": "A", "content": "$$[{M^{ - 3}}{L^{ - 4}}{I^3}{T^7}]$$"}, {"identifier": "B", "content": "$$[{M^{ - 2}}{L^{ - 3}}{I^2}{T^6}]$$"}, {"identifier": "C", "content": "$$[{M^{ - 2}}{L^{ - 4}}{I^3}{T^7}]$$"}, {"identifier": "D", "content": "$$[{M^{ - 1}}{L^{ - 3}}{I^{ - 2}}{T^{ - 7}}]$$"}] | ["C"] | null | $$\lambda = {C \over V} = {{Q/V} \over V} = {Q \over {{V^2}}}$$<br><br>$$V = {{work} \over Q}$$<br><br>$$\lambda = {{{Q^3}} \over {{{(work)}^2}}} = {{{{(It)}^3}} \over {{{(F.s)}^2}}}$$<br><br>$$ = {{\left[ {{I^3}{T^3}} \right]} \over {{{\left[ {M{L^2}{T^{ - 2}}} \right]}^2}}} = [{M^{ - 2}}{L^{ - 4}}{I^3}{T^7}]$$ | mcq | jee-main-2021-online-26th-february-evening-slot | 12,823 |
1krqco49l | physics | units-and-measurements | dimensions-of-physical-quantities | If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is : | [{"identifier": "A", "content": "$$[{t^{ - 1}}{v^1}{l^{ - 2}}]$$"}, {"identifier": "B", "content": "$$[{t^1}{v^2}{l^{ - 1}}]$$"}, {"identifier": "C", "content": "$$[{t^{ - 2}}{v^{ - 1}}{l^1}]$$"}, {"identifier": "D", "content": "$$[{t^{ - 1}}{v^{ - 2}}{l^1}]$$"}] | ["D"] | null | $$m \propto {t^a}{v^b}{l^c}$$<br><br>$$m \propto {[T]^a}{[L{T^{ - 1}}]^b}{[M{L^2}{T^{ - 1}}]^c}$$<br><br>$${M^1}{L^0}{T^0} = {M^c}{L^{b + 2c}}{T^{a - b - c}}$$<br><br>comparing powers<br><br>c = 1, b = $$-$$2, a = $$-$$1<br><br>$$m \propto {t^{ - 1}}{v^{ - 2}}{l^1}$$ | mcq | jee-main-2021-online-20th-july-evening-shift | 12,824 |
1krw9mhlm | physics | units-and-measurements | dimensions-of-physical-quantities | The force is given in terms of time t and displacement x by the equation <br/><br/>F = A cos Bx + C sin Dt<br/><br/>The dimensional formula of $${{AD} \over B}$$ is : | [{"identifier": "A", "content": "$$[{M^0}L{T^{ - 1}}]$$"}, {"identifier": "B", "content": "$$[M{L^2}{T^{ - 3}}]$$"}, {"identifier": "C", "content": "$$[{M^1}{L^1}{T^{ - 2}}]$$"}, {"identifier": "D", "content": "$$[{M^2}{L^2}{T^{ - 3}}]$$"}] | ["B"] | null | $$[A] = [ML{T^{ - 2}}]$$<br><br>$$[B] = [{L^{ - 1}}]$$<br><br>$$[D] = [{T^{ - 1}}]$$<br><br>$$\left[ {{{AD} \over B}} \right] = {{[ML{T^{ - 2}}][{T^{ - 1}}]} \over {[{L^{ - 1}}]}}$$<br><br>$$\left[ {{{AD} \over B}} \right] = [M{L^2}{T^{ - 3}}]$$ | mcq | jee-main-2021-online-25th-july-evening-shift | 12,825 |
1ktafapj6 | physics | units-and-measurements | dimensions-of-physical-quantities | If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL<sup>2</sup>M<sup>$$-$$5</sup>G<sup>$$-$$2</sup> are : | [{"identifier": "A", "content": "[M<sup>0</sup> L<sup>1</sup> T<sup>0</sup>]"}, {"identifier": "B", "content": "[M<sup>$$-$$1</sup> L<sup>$$-$$1</sup> T<sup>2</sup>]"}, {"identifier": "C", "content": "[M<sup>1</sup> L<sup>1</sup> T<sup>$$-$$2</sup>]"}, {"identifier": "D", "content": "[M<sup>0</sup> L<sup>0</sup> T<sup>... | ["D"] | null | E = ML<sup>2</sup>T<sup>$$-$$2</sup><br><br>L = ML<sup>2</sup>T<sup>$$-$$1</sup><br><br>m = M<br><br>G = M<sup>$$-$$1</sup>L<sup>+3</sup>T<sup>$$-$$2</sup><br><br>P = $${{E{L^2}} \over {{M^5}{G^2}}}$$<br><br>[P] = $${{(M{L^2}{T^{ - 2}})({M^2}{L^4}{T^{ - 2}})} \over {{M^5}({M^{ - 2}}{L^6}{T^{ - 4}})}} = {M^0}{L^0}{T^0}$... | mcq | jee-main-2021-online-26th-august-morning-shift | 12,826 |
1ktbraocr | physics | units-and-measurements | dimensions-of-physical-quantities | Match List - I with List - II<br/><br/> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267766/exam_images/bhydvvpqezynvzrnhoq0.webp"/><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265881/exam_images/xivv1fvf7... | [{"identifier": "A", "content": "(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)"}, {"identifier": "B", "content": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)"}, {"identifier": "C", "content": "(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)"}, {"identifier": "D", "content": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)"}] | ["D"] | null | (a) Magnetic Induction = MT<sup>$$-$$2</sup>A<sup>$$-$$1</sup><br><br>(b) Magnetic Flux = ML<sup>2</sup>T<sup>$$-$$2</sup>A<sup>$$-$$1</sup><br><br>(c) Magnetic Permeability = MLT<sup>$$-$$2</sup>A<sup>$$-$$2</sup><br><br>(d) Magnetization = M<sup>0</sup>L<sup>$$-$$1</sup>A | mcq | jee-main-2021-online-26th-august-evening-shift | 12,827 |
1ktdyiybr | physics | units-and-measurements | dimensions-of-physical-quantities | Which of the following is not a dimensionless quantity? | [{"identifier": "A", "content": "Relative magnetic permeability ($$\\mu$$<sub>r</sub>)"}, {"identifier": "B", "content": "Power factor"}, {"identifier": "C", "content": "Permeability of free space ($$\\mu$$<sub>0</sub>)"}, {"identifier": "D", "content": "Quality factor"}] | ["C"] | null | [$$\mu$$<sub>r</sub>] = 1 as $$\mu$$<sub>r</sub> = $${\mu \over {{\mu _m}}}$$<br><br>[power factor (cos $$\phi$$)] = 1<br><br>$${\mu _0} = {{{B_0}} \over H}$$ (unit = NA<sup>$$-$$2</sup>) : Not dimensionless<br><br>[$$\mu$$<sub>0</sub>] = [MLT<sup>$$-$$2</sup>A<sup>$$-$$2</sup>]<br><br>quality factor $$(Q) = {{Energy\... | mcq | jee-main-2021-online-27th-august-morning-shift | 12,828 |
1ktfmwreu | physics | units-and-measurements | dimensions-of-physical-quantities | If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density : | [{"identifier": "A", "content": "[FL<sup>$$-$$4</sup>T<sup>2</sup>]"}, {"identifier": "B", "content": "[FL<sup>$$-$$3</sup>T<sup>2</sup>]"}, {"identifier": "C", "content": "[FL<sup>$$-$$5</sup>T<sup>2</sup>]"}, {"identifier": "D", "content": "[FL<sup>$$-$$3</sup>T<sup>3</sup>]"}] | ["A"] | null | Density = [F<sup>a</sup>L<sup>b</sup>T<sup>c</sup>]<br><br>[ML<sup>$$-$$3</sup>] = [M<sup>a</sup>L<sup>a+b</sup>T<sup>$$-$$2a</sup>L<sup>b</sup>T<sup>c</sup>]<br><br>[M<sup>1</sup>L<sup>$$-$$3</sup>] = [M<sup>a</sup>L<sup>a+b</sup>T<sup>$$-$$2a+c</sup>]<br><br>$$\matrix{
{a = 1} & ; & {a + b = - 3} & ; ... | mcq | jee-main-2021-online-27th-august-evening-shift | 12,829 |
1kth4i2zc | physics | units-and-measurements | dimensions-of-physical-quantities | Match List - I with List - II.<br/><br/><table>
<thead>
<tr>
<th></th>
<th>List - I</th>
<th></th>
<th>List - II</th>
</tr>
</thead>
<tbody>
<tr>
<td>(a)</td>
<td>Torque</td>
<td>(i)</td>
<td>MLT$$^{ - 1}$$</td>
</tr>
<tr>
<td>(b)</td>
<td>Impulse</td>
<td>(ii)</td>
<td>MT$$^{ - 2}$$</td>
</tr>
<tr>
<td>(c)</td>
<td>Te... | [{"identifier": "A", "content": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)"}, {"identifier": "B", "content": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)"}, {"identifier": "C", "content": "(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)"}, {"identifier": "D", "content": "(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)"}] | ["A"] | null | torque $$\tau$$ $$\to$$ ML<sup>2</sup>T<sup>$$-$$2</sup> (iii)<br><br>Impulse I $$\Rightarrow$$ MLT<sup>$$-$$1</sup> (i)<br><br>Tension force $$\Rightarrow$$ MLT<sup>$$-$$2</sup> (iv)<br><br>Surface tension $$\Rightarrow$$ MT<sup>$$-$$2</sup> (ii)<br><br>Option (a) | mcq | jee-main-2021-online-31st-august-morning-shift | 12,830 |
1kth50gxo | physics | units-and-measurements | dimensions-of-physical-quantities | Which of the following equations is dimensionally incorrect?<br/><br/>Where t = time, h = height, s = surface tension, $$\theta$$ = angle, $$\rho$$ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, T = torque, $$\in$$ = permittivity, E = electric field, J = current dens... | [{"identifier": "A", "content": "$$v = {{\\pi p{a^4}} \\over {8\\eta L}}$$"}, {"identifier": "B", "content": "$$h = {{2s\\cos \\theta } \\over {\\rho rg}}$$"}, {"identifier": "C", "content": "$$J = \\in {{\\partial E} \\over {\\partial t}}$$"}, {"identifier": "D", "content": "$$W = \\Gamma \\theta $$"}] | ["A"] | null | (a) $${{\pi p{a^4}} \over {8\eta L}} = {{dv} \over {dt}}$$ = Volumetric flow rate (Poiseuille's law)<br><br>(b) $$h\rho g = {{2s} \over r}\cos \theta $$<br><br>(c) $$\varepsilon \times {1 \over {4\pi {\varepsilon _0}}}{a \over {{r^2}}} \times {1 \over \varepsilon } = {q \over t} \times {1 \over {{r^2}}}$$<br><br>$$ = ... | mcq | jee-main-2021-online-31st-august-morning-shift | 12,831 |
1ktjoqnm5 | physics | units-and-measurements | dimensions-of-physical-quantities | If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be : | [{"identifier": "A", "content": "[FT<sup>$$-$$1</sup> V<sup>$$-$$1</sup>]"}, {"identifier": "B", "content": "[FTV<sup>$$-$$1</sup>]"}, {"identifier": "C", "content": "[FT<sup>2</sup> V]"}, {"identifier": "D", "content": "[FVT<sup>$$-$$1</sup>]"}] | ["B"] | null | [M] = K[F]<sup>a</sup> [T]<sup>b</sup> [V]<sup>c</sup><br><br>[M<sup>1</sup>] = [M<sup>1</sup>L<sup>1</sup>T<sup>$$-$$2</sup>]<sup>a</sup> [T<sup>1</sup>]<sup>b</sup> [L<sup>1</sup>T<sup>$$-$$1</sup>]<sup>c</sup><br><br>a = 1, b = 1, c = $$-$$1<br><br>$$\therefore$$ [M] = [FTV<sup>$$-$$1</sup>] | mcq | jee-main-2021-online-31st-august-evening-shift | 12,832 |
1l56ucya7 | physics | units-and-measurements | dimensions-of-physical-quantities | <p>The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be :</p> | [{"identifier": "A", "content": "[ML<sup>$$-$$1</sup>T<sup>$$-$$1</sup>]"}, {"identifier": "B", "content": "[ML<sup>$$-$$1</sup>T<sup>$$-$$2</sup>]"}, {"identifier": "C", "content": "[ML<sup>2</sup>T<sup>$$-$$1</sup>]"}, {"identifier": "D", "content": "[M<sup>$$-$$1</sup>L<sup>3</sup>T<sup>0</sup>]"}] | ["A"] | null | <p>[pascal-second] = $${{ML{T^{ - 2}}} \over {{L^2}}} \times T$$</p>
<p>$$ = M{L^{ - 1}}{T^{ - 1}}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 12,833 |
1l58b9vrj | physics | units-and-measurements | dimensions-of-physical-quantities | <p>An expression for a dimensionless quantity P is given by $$P = {\alpha \over \beta }{\log _e}\left( {{{kt} \over {\beta x}}} \right)$$; where $$\alpha$$ and $$\beta$$ are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of $$\alpha$$ will be :</p> | [{"identifier": "A", "content": "[M<sup>0</sup> L<sup>$$-$$1</sup> T<sup>0</sup>]"}, {"identifier": "B", "content": "[M L<sup>0</sup> T<sup>$$-$$2</sup>]"}, {"identifier": "C", "content": "[M L T<sup>$$-$$2</sup>]"}, {"identifier": "D", "content": "[M L<sup>2</sup> T<sup>$$-$$2</sup>]"}] | ["C"] | null | Given, $P=\frac{\alpha}{\beta} \log _{e}\left[\frac{k t}{\beta x}\right]$
<br/><br/>The logarithmic term is dimensionless.
<br/><br/>Thus, $[k t / \beta x]$ is also dimensionless.
<br/><br/>i.e. $\frac{[k][t]}{[\beta][x]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$ .......(i)
<br/><br/>We have, $E=... | mcq | jee-main-2022-online-26th-june-morning-shift | 12,834 |
1l58h9trw | physics | units-and-measurements | dimensions-of-physical-quantities | <p>The dimension of mutual inductance is :</p> | [{"identifier": "A", "content": "$$[M{L^2}{T^{ - 2}}{A^{ - 1}}]$$"}, {"identifier": "B", "content": "$$[M{L^2}{T^{ - 3}}{A^{ - 1}}]$$"}, {"identifier": "C", "content": "$$[M{L^2}{T^{ - 2}}{A^{ - 2}}]$$"}, {"identifier": "D", "content": "$$[M{L^2}{T^{ - 3}}{A^{ - 2}}]$$"}] | ["C"] | null | <p>$$\because$$ $$U = {1 \over 2}M{i^2}$$</p>
<p>$$ \Rightarrow [M] = {{[U]} \over {[{i^2}]}} = {{M{L^2}{T^{ - 2}}} \over {{A^2}}}$$</p>
<p>$$ = [M{L^2}{T^{ - 2}}{A^{ - 2}}]$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 12,835 |
1l5bbdwo9 | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Identify the pair of physical quantities that have same dimensions:</p> | [{"identifier": "A", "content": "velocity gradient and decay constant"}, {"identifier": "B", "content": "Wien's constant and Stefan constant"}, {"identifier": "C", "content": "angular frequency and angular momentum"}, {"identifier": "D", "content": "wave number and Avogadro number"}] | ["A"] | null | <p>Velocity gradient $$ = {{dv} \over {dx}}$$</p>
<p>$$\Rightarrow$$ Dimensions are $${{[L{T^{ - 1}}]} \over {[L]}} = [{T^{ - 1}}]$$</p>
<p>Decay constant $$\lambda$$ has dimensions of $$[{T^{ - 1}}]$$ because of the relation $${{dN} \over {dt}} = - \lambda $$ N</p>
<p>$$\Rightarrow$$ Velocity gradient and decay const... | mcq | jee-main-2022-online-24th-june-evening-shift | 12,836 |
1l5c2rikj | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Identify the pair of physical quantities which have different dimensions:</p> | [{"identifier": "A", "content": "Wave number and Rydberg's constant"}, {"identifier": "B", "content": "Stress and Coefficient of elasticity"}, {"identifier": "C", "content": "Coercivity and Magnetisation"}, {"identifier": "D", "content": "Specific heat capacity and Latent heat"}] | ["D"] | null | <p>$$[S] = {{[C]} \over {[m] \times [\Delta T]}}$$</p>
<p>and, $$[L] = {{[Q]} \over {[m]}}$$</p>
<p>$$\Rightarrow$$ They have different dimensions.</p> | mcq | jee-main-2022-online-24th-june-morning-shift | 12,837 |
1l6km9t7o | physics | units-and-measurements | dimensions-of-physical-quantities | <p>An expression of energy density is given by $$u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$$, where $$\alpha, \beta$$ are constants, $$x$$ is displacement, $$k$$ is Boltzmann constant and t is the temperature. The dimensions of $$\beta$$ will be :</p> | [{"identifier": "A", "content": "$$\\left[\\mathrm{ML}^{2} \\mathrm{~T}^{-2} \\theta^{-1}\\right]$$"}, {"identifier": "B", "content": "$$\\left[\\mathrm{M}^{0} \\mathrm{~L}^{2} \\mathrm{~T}^{-2}\\right]$$"}, {"identifier": "C", "content": "$$\\left[\\mathrm{M}^{0} \\mathrm{~L}^{0} \\mathrm{~T}^{0}\\right]$$"}, {"identi... | ["D"] | null | <p>$$u = {\alpha \over \beta }\sin \left( {{{\alpha x} \over {kt}}} \right)$$</p>
<p>$$[\alpha] = \left[ {{{kt} \over x}} \right] = {{[Energy]} \over {[Dis\tan ce]}}$$</p>
<p>$$[\beta ] = {{[\alpha ]} \over {[u]}}$$</p>
<p>$$ = {{[Energy]/[Dis\tan ce]} \over {[Energy]/[Volume]}}$$</p>
<p>$$ = [{L^2}]$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift | 12,839 |
1l6npvqpp | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Consider the efficiency of carnot's engine is given by $$\eta=\frac{\alpha \beta}{\sin \theta} \log_e \frac{\beta x}{k T}$$, where $$\alpha$$ and $$\beta$$ are constants. If T is temperature, k is Boltzmann constant, $$\theta$$ is angular displacement and x has the dimensions of length. Then, choose the incorrect op... | [{"identifier": "A", "content": "Dimensions of $$\\beta$$ is same as that of force."}, {"identifier": "B", "content": "Dimensions of $$\\alpha^{-1} x$$ is same as that of energy."}, {"identifier": "C", "content": "Dimensions of $$\\eta^{-1} \\sin \\theta$$ is same as that of $$\\alpha \\beta$$."}, {"identifier": "D", "... | ["D"] | null | Since, dimensions trigonometric function and logarithmic function are dimensionless quantities.
<br/><br/>$$
\therefore[\eta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]
<br/><br/>$$
<br/><br/>Also, dimensions of temperature, $[T]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}{ }^0 \mathrm{~K}\right]$
<br/><br/... | mcq | jee-main-2022-online-28th-july-evening-shift | 12,841 |
1l6p3y0nm | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).</p>
<p>Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is $$\rho$$ and radius of the drop is r, then $$\mathrm{T}=\mathrm{K} \sqrt{\rho \mathrm{r... | [{"identifier": "A", "content": "Both (A) and (R) are true and (R) is the correct explanation of (A)"}, {"identifier": "B", "content": "Both (A) and (R) are true but (R) is not the correct explanation of (A)"}, {"identifier": "C", "content": "(A) is true but (R) is false"}, {"identifier": "D", "content": "(A) is false ... | ["D"] | null | We know,
<br/><br/>$$
\begin{gathered}
{[S]=\left[\mathrm{MT}^{-2}\right]} \\\\
{[\rho]=\left[\mathrm{ML}^{-3}\right]} \\\\
{[r]=\left[\mathrm{L}]\right.}
\end{gathered}
$$
<br/><br/>$$
\begin{aligned}
\therefore \text { Dimension of } \mathrm{RHS} &=\frac{\left[\mathrm{M}^{\frac{1}{2}} \mathrm{~L}^{-\frac{3}{2}}\right... | mcq | jee-main-2022-online-29th-july-morning-shift | 12,842 |
ldo74ikz | physics | units-and-measurements | dimensions-of-physical-quantities | Match List <b>I</b> with List <b>II</b><br/><br/>
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-... | [{"identifier": "A", "content": "A - I, B - IV, C - III, D - II"}, {"identifier": "B", "content": "A - III, B - I, C - IV, D - II"}, {"identifier": "C", "content": "A - IV, B - II, C - I, D - III"}, {"identifier": "D", "content": "A - II, B - III, C - IV, D - I"}] | ["B"] | null | $\vec{L}=\vec{r} \times \vec{p} \Rightarrow[\mathrm{L}]=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]$
<br/><br/>$$
=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]
$$
<br/><br/>$$
\begin{aligned}
\vec{\tau}=\vec{r} \times \vec{F} ... | mcq | jee-main-2023-online-31st-january-evening-shift | 12,844 |
1lds9liqj | physics | units-and-measurements | dimensions-of-physical-quantities | <p>The equation of a circle is given by $$x^2+y^2=a^2$$, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : $${(x - At)^2} + {\left( {y - {t \over B}} \right)^2} = {a^2}$$. The dimensions of t is given as $$[\mat... | [{"identifier": "A", "content": "$$\\mathrm{A=[L^{-1}T^{-1}],B=[LT^{-1}]}$$"}, {"identifier": "B", "content": "$$\\mathrm{A=[L^{-1}T^{-1}],B=[LT]}$$"}, {"identifier": "C", "content": "$$\\mathrm{A=[LT],B=[L^{-1}T^{-1}]}$$"}, {"identifier": "D", "content": "$$\\mathrm{A=[L^{-1}T],B=[LT^{-1}]}$$"}] | ["C"] | null | <p>Here, At is distance, so dimensions of</p>
<p>$$[At] = [x] = [L]$$</p>
<p>Given. The dimensions of t is $$[\mathrm{T^{-1}]}$$</p>
<p>${\left[A \times \mathrm{T}^{-1}\right]=[\mathrm{L}] \Rightarrow[A]=[\mathrm{LT}]}$</p>
<p>$$\left[ {{t \over B}} \right] = [y] = [L]$$</p>
<p>$\Rightarrow \frac{\mathrm{T}^{-1}}{[B]}=... | mcq | jee-main-2023-online-29th-january-evening-shift | 12,845 |
1ldsp679n | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Match List I with List II :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;... | [{"identifier": "A", "content": "A-III, B-II, C-IV, D-I"}, {"identifier": "B", "content": "A-III, B-II, C-I, D-IV"}, {"identifier": "C", "content": "A-II, B-III, C-IV, D-I"}, {"identifier": "D", "content": "A-II, B-III, C-I, D-IV"}] | ["A"] | null | Pressure gradient $=\frac{d p}{d x}=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{[\mathrm{L}]}$
<br/><br/>
$$
=\left[\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]
$$
<br/><br/>
Energy density $=\frac{\text { energy }}{\text { volume }}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathr... | mcq | jee-main-2023-online-29th-january-morning-shift | 12,846 |
1ldtyxe4z | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-II, B-III, C-IV, D-I"}, {"identifier": "B", "content": "A-I, B-II, C-III, D-IV"}, {"identifier": "C", "content": "A-I, B-III, C-IV, D-II"}, {"identifier": "D", "content": "A-III, B-I, C-II, D-IV"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\\\
& \mathrm{F}=6 \pi \eta \mathrm{rv} \Rightarrow \eta=\frac{\mathrm{F}}{6 \pi \... | mcq | jee-main-2023-online-25th-january-evening-shift | 12,847 |
1ldyef9zw | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-I, B-III, C-IV, D-II"}, {"identifier": "B", "content": "A-III, B-IV, C-I, D-II"}, {"identifier": "C", "content": "A-II, B-IV, C-III, D-I"}, {"identifier": "D", "content": "A-III, B-I, C-II, D-IV"}] | ["B"] | null | (A) Planck's constant<br/><br/>
$$
\begin{aligned}
& \mathrm{h} v=\mathrm{E} \\\\
& \mathrm{h}=\frac{\mathrm{E}}{v}=\frac{\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1} \quad...(\text{III})
\end{aligned}
$$<br/><br/>
(B) $\mathrm{E}=\mathrm{qV}$<br/><br/>
$$
V... | mcq | jee-main-2023-online-24th-january-morning-shift | 12,848 |
1lgoyia4x | physics | units-and-measurements | dimensions-of-physical-quantities | <p>In the equation $$\left[X+\frac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T, X$$ is pressure, $$Y$$ is volume, $$\mathrm{R}$$ is universal gas constant and $$T$$ is temperature. The physical quantity equivalent to the ratio $$\frac{a}{b}$$ is:</p> | [{"identifier": "A", "content": "Impulse"}, {"identifier": "B", "content": "Energy"}, {"identifier": "C", "content": "Pressure gradient"}, {"identifier": "D", "content": "Coefficient of viscosity"}] | ["B"] | null | Given that, $$\left[X+\frac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T$$<br/><br/>$$ \therefore $$ $X$ and $\frac{a}{Y^2}$ have the same dimensions and $Y$ and $b$ have the same dimensions, let's analyze the dimensions of $\frac{a}{b}$.
<br/><br/>
Since $X$ represents pressure, it has dimensions of $[M L^{-1} T^{-2}]$.
<br/><... | mcq | jee-main-2023-online-13th-april-evening-shift | 12,850 |
1lh00yauq | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Dimension of $$\frac{1}{\mu_{0} \in_{0}}$$ should be equal to</p> | [{"identifier": "A", "content": "$$\\mathrm{T}^{2} / \\mathrm{L}^{2}$$"}, {"identifier": "B", "content": "$$\\mathrm{T} / \\mathrm{L}$$"}, {"identifier": "C", "content": "$$\\mathrm{L}^{2} / \\mathrm{T}^{2}$$"}, {"identifier": "D", "content": "$$\\mathrm{L} / \\mathrm{T}$$"}] | ["C"] | null | The term $\frac{1}{\mu_{0} \epsilon_{0}}$ appears in the formula for the speed of light $c$, which is:
<br/><br/>
$c = \sqrt{\frac{1}{\mu_{0} \epsilon_{0}}}$
<br/><br/>
where $\mu_{0}$ is the permeability of free space and $\epsilon_{0}$ is the permittivity of free space.
<br/><br/>
The speed of light $c$ has dimensio... | mcq | jee-main-2023-online-8th-april-morning-shift | 12,852 |
lsbknus1 | physics | units-and-measurements | dimensions-of-physical-quantities | The dimensional formula of angular impulse is : | [{"identifier": "A", "content": "$\\left[\\mathrm{M} \\mathrm{L}^2 \\mathrm{~T}^{-2}\\right]$"}, {"identifier": "B", "content": "$\\left[\\mathrm{M} \\mathrm{L}^{-2} \\mathrm{~T}^{-1}\\right]$"}, {"identifier": "C", "content": "$\\left[\\mathrm{M} \\mathrm{L}^2 \\mathrm{~T}^{-1}\\right]$"}, {"identifier": "D", "content... | ["C"] | null | <p>Angular impulse is given when a torque is applied for a certain amount of time. The angular impulse changes the angular momentum of an object and has the same dimensions as angular momentum.</p>
<p>The dimensional formula for torque $\tau$ is the same as that for work, since torque is a kind of rotational work, whic... | mcq | jee-main-2024-online-1st-february-morning-shift | 12,853 |
jaoe38c1lsc3tnnj | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Given below are two statements :</p>
<p>Statement (I) : Planck's constant and angular momentum have same dimensions.</p>
<p>Statement (II) : Linear momentum and moment of force have same dimensions.</p>
<p>In the light of the above statements, choose the correct answer from the options given below :</p> | [{"identifier": "A", "content": "Statement I is true but Statement II is false\n"}, {"identifier": "B", "content": "Both Statement I and Statement II are false\n"}, {"identifier": "C", "content": "Both Statement I and Statement II are true\n"}, {"identifier": "D", "content": "Statement I is false but Statement II is tr... | ["A"] | null | <p>$$\begin{aligned}
& {[\mathrm{h}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\
& {[\mathrm{~L}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\
& {[\mathrm{P}]=\mathrm{MLT}^{-1}} \\
& {[\tau]=\mathrm{ML}^2 \mathrm{~T}^{-2}}
\end{aligned}$$</p>
<p>(Here $$\mathrm{h}$$ is Planck's constant, $$\mathrm{L}$$ is angular momentum, $$\mathrm{P}$$... | mcq | jee-main-2024-online-27th-january-morning-shift | 12,854 |
jaoe38c1lsd5ibjj | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Consider two physical quantities $$A$$ and $$B$$ related to each other as $$E=\frac{B-x^2}{A t}$$ where $$E, x$$ and $$t$$ have dimensions of energy, length and time respectively. The dimension of $$A B$$ is</p> | [{"identifier": "A", "content": "$$L^{-2} M^1 T^0$$\n"}, {"identifier": "B", "content": "$$L^2 M^{-1} T^1$$\n"}, {"identifier": "C", "content": "$$L^0 M^{-1} T^1$$\n"}, {"identifier": "D", "content": "$$L^{-2} M^{-1} T^1$$"}] | ["B"] | null | <p>$$\begin{aligned}
& {[\mathrm{B}]=\mathrm{L}^2} \\
& \mathrm{~A}=\frac{\mathrm{x}^2}{\mathrm{tE}}=\frac{\mathrm{L}^2}{\mathrm{TML}^2 \mathrm{~T}^{-2}}=\frac{1}{\mathrm{MT}^{-1}} \\
& {[\mathrm{~A}]=\mathrm{M}^{-1 \mathrm{~T}}} \\
& {[\mathrm{AB}]=\left[\mathrm{L}^2 \mathrm{M}^{-1} \mathrm{~T}^1\right]}
\end{aligned}... | mcq | jee-main-2024-online-31st-january-evening-shift | 12,855 |
jaoe38c1lse6mdlu | physics | units-and-measurements | dimensions-of-physical-quantities | <p>A force is represented by $$F=a x^2+b t^{\frac{1}{2}}$$</p>
<p>where $$x=$$ distance and $$t=$$ time. The dimensions of $$b^2 / a$$ are:</p> | [{"identifier": "A", "content": "$$\\left[\\mathrm{ML}^2 \\mathrm{~T}^{-3}\\right]$$\n"}, {"identifier": "B", "content": "$$\\left[\\mathrm{ML}^3 \\mathrm{~T}^{-3}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[M L T^{-2}\\right]$$\n"}, {"identifier": "D", "content": "$$\\left[M L^{-1} T^{-1}\\right]$$"}] | ["B"] | null | <p>To determine the dimensions of $$\frac{b^2}{a}$$, let's start by identifying the dimensions of each term in the equation $$F=a x^2+b t^{\frac{1}{2}}$$, where $$F$$ represents force, $$x$$ represents distance, and $$t$$ represents time.</p>
<p>The dimension of force ($$F$$) is given by [MLT<sup>-2</sup>], where $$M$$... | mcq | jee-main-2024-online-31st-january-morning-shift | 12,856 |
1lsg6udgl | physics | units-and-measurements | dimensions-of-physical-quantities | <p>If mass is written as $$m=k \mathrm{c}^{\mathrm{P}} G^{-1 / 2} h^{1 / 2}$$ then the value of $$P$$ will be : (Constants have their usual meaning with $k a$ dimensionless constant)</p> | [{"identifier": "A", "content": "1/3"}, {"identifier": "B", "content": "$$-$$1/3"}, {"identifier": "C", "content": "1/2"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>$$\begin{aligned}
& \mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2} \\
& \mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2}
\end{aligned}... | mcq | jee-main-2024-online-30th-january-evening-shift | 12,857 |
1lsgdjves | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Match List I with List II.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;b... | [{"identifier": "A", "content": "(A)-(II), (B)-(I), (C)-(IV), (D)-(III)\n"}, {"identifier": "B", "content": "(A)-(I), (B)-(II), (C)-(III), (D)-(IV)\n"}, {"identifier": "C", "content": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)\n"}, {"identifier": "D", "content": "(A)-(III), (B)-(IV), (C)-(II), (D)-(I)"}] | ["D"] | null | <p>$$\begin{aligned}
& F=\eta A \frac{d v}{d y} \\
& {\left[M L T^{-2}\right]=\eta\left[L^2\right]\left[T^{-1}\right]} \\
& \eta=\left[M L^{-1} T^{-1}\right] \\
& S . T=\frac{F}{\ell}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^0 T^{-2}\right] \\
& L=m v r=\left[M L^2 T^{-1}\right] \\
& K . E=\frac{1}{2} I \omega^2=\... | mcq | jee-main-2024-online-30th-january-morning-shift | 12,858 |
luxwd2ms | physics | units-and-measurements | dimensions-of-physical-quantities | <p>The de-Broglie wavelength associated with a particle of mass $$m$$ and energy $$E$$ is $$h / \sqrt{2 m E}$$. The dimensional formula for Planck's constant is :</p> | [{"identifier": "A", "content": "$$\\left[\\mathrm{M}^2 \\mathrm{~L}^2 \\mathrm{~T}^{-2}\\right]$$\n"}, {"identifier": "B", "content": "$$\\left[\\mathrm{ML}^{-1} \\mathrm{~T}^{-2}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[\\mathrm{ML}^2 \\mathrm{~T}^{-1}\\right]$$\n"}, {"identifier": "D", "content": "$$\... | ["C"] | null | <p>To determine the dimensional formula for Planck's constant, we will start by analyzing the given de-Broglie wavelength equation:</p>
<p>$$\lambda = \frac{h}{\sqrt{2 m E}}$$</p>
<p>Here, $$\lambda$$ is the wavelength, $$h$$ is the Planck's constant, $$m$$ is the mass of the particle, and $$E$$ is the energy of the ... | mcq | jee-main-2024-online-9th-april-evening-shift | 12,859 |
luy9clxz | physics | units-and-measurements | dimensions-of-physical-quantities | <p>The dimensional formula of latent heat is :</p> | [{"identifier": "A", "content": "$$\\left[\\mathrm{M}^{\\mathrm{0}} \\mathrm{LT}^{-2}\\right]$$\n"}, {"identifier": "B", "content": "$$\\left[\\mathrm{MLT}^{-2}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[\\mathrm{M}^0 \\mathrm{~L}^2 \\mathrm{~T}^{-2}\\right]$$\n"}, {"identifier": "D", "content": "$$\\left[... | ["C"] | null | <p>To derive the dimensional formula of latent heat, we need to understand what latent heat actually refers to. Latent heat is the amount of heat absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.</p>
<p>The formula for latent heat ($L$) is g... | mcq | jee-main-2024-online-9th-april-morning-shift | 12,860 |
lv0vxm6r | physics | units-and-measurements | dimensions-of-physical-quantities | <p>The equation of stationary wave is :</p>
<p>$$y=2 \mathrm{a} \sin \left(\frac{2 \pi \mathrm{nt}}{\lambda}\right) \cos \left(\frac{2 \pi x}{\lambda}\right) \text {. }$$</p>
<p>Which of the following is NOT correct :</p> | [{"identifier": "A", "content": "The dimensions of $$\\mathrm{nt}$$ is [L]\n"}, {"identifier": "B", "content": "The dimensions of $$n$$ is $$[\\mathrm{LT}^{-1}]$$\n"}, {"identifier": "C", "content": "The dimensions of $$x$$ is [L]\n"}, {"identifier": "D", "content": "The dimensions of $$n / \\lambda$$ is [T]"}] | ["D"] | null | <p>To determine which of the options is NOT correct, we need to analyze the dimensional consistency of each term in the given equation of the stationary wave:</p>
<p>$$y = 2a \sin \left(\frac{2\pi nt}{\lambda}\right) \cos \left(\frac{2\pi x}{\lambda}\right).$$</p>
<p>Let's break down the dimensions for each relevant ... | mcq | jee-main-2024-online-4th-april-morning-shift | 12,861 |
lv2es2kt | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Applying the principle of homogeneity of dimensions, determine which one is correct,
where $$T$$ is time period, $$G$$ is gravitational constant, $$M$$ is mass, $$r$$ is radius of orbit.</p> | [{"identifier": "A", "content": "$$T^2=\\frac{4 \\pi^2 r}{G M^2}$$\n"}, {"identifier": "B", "content": "$$T^2=4 \\pi^2 r^3$$\n"}, {"identifier": "C", "content": "$$T^2=\\frac{4 \\pi^2 r^3}{G M}$$\n"}, {"identifier": "D", "content": "$$T^2=\\frac{4 \\pi^2 r^2}{G M}$$"}] | ["C"] | null | <p>To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions. The time period ($$T$$) is measured in units of time ($$T$$), the gravitational constant ($$G$$) has units of $$\text{m}^3\text{kg}^{-1}\text{s}^{-2}$$... | mcq | jee-main-2024-online-4th-april-evening-shift | 12,862 |
lv3ve84f | physics | units-and-measurements | dimensions-of-physical-quantities | <p>If $$\epsilon_{\mathrm{o}}$$ is the permittivity of free space and $$\mathrm{E}$$ is the electric field, then $$\epsilon_{\mathrm{o}} \mathrm{E}^2$$ has the dimensions :</p> | [{"identifier": "A", "content": "$$[\\mathrm{M} \\mathrm{L}^{-1} \\mathrm{~T}^{-2}]$$\n"}, {"identifier": "B", "content": "$$[\\mathrm{M} \\mathrm{L}^2 \\mathrm{~T}^{-2}]$$\n"}, {"identifier": "C", "content": "$$[\\mathrm{M}^{\\mathrm{0}} \\mathrm{L}^{-2} \\mathrm{TA}]$$\n"}, {"identifier": "D", "content": "$$[\\mathrm... | ["A"] | null | <p>To determine the dimensions of $$\epsilon_{\mathrm{o}} \mathrm{E}^2$$, we need to first understand the dimensional formulas of each component in the expression.</p>
<p>1. <strong>Permittivity of free space, $$\epsilon_{\mathrm{o}}$$:</strong> </p>
<p>The permittivity of free space has the dimensions: $$\left[\epsi... | mcq | jee-main-2024-online-8th-april-evening-shift | 12,863 |
lv9s20rh | physics | units-and-measurements | dimensions-of-physical-quantities | <p>What is the dimensional formula of $$a b^{-1}$$ in the equation $$\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$$, where letters have their usual meaning.</p> | [{"identifier": "A", "content": "$$[\\mathrm{M}^6 \\mathrm{~L}^7 \\mathrm{~T}^4]$$\n"}, {"identifier": "B", "content": "$$[\\mathrm{M}^{-1} \\mathrm{~L}^5 \\mathrm{~T}^3]$$\n"}, {"identifier": "C", "content": "$$[\\mathrm{M}^0 \\mathrm{~L}^3 \\mathrm{~T}^{-2}]$$\n"}, {"identifier": "D", "content": "$$[\\mathrm{ML}^2 \\... | ["D"] | null | <p>To find the dimensional formula of $$a b^{-1}$$ in the equation given by $$\left( P + \frac{a}{V^2}\right)(V-b) = RT$$, where $P$ is the pressure, $V$ is the volume, and $T$ is the temperature, we will first understand the dimensions of each term in the equation. The variables mentioned have their usual meanings in ... | mcq | jee-main-2024-online-5th-april-evening-shift | 12,864 |
lvc57i1o | physics | units-and-measurements | dimensions-of-physical-quantities | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-I, B-III, C-II, D-IV\n"}, {"identifier": "B", "content": "A-IV, B-II, C-III, D-I\n"}, {"identifier": "C", "content": "A-III, B-I, C-II, D-IV\n"}, {"identifier": "D", "content": "A-IV, B-III, C-II, D-I"}] | ["D"] | null | <p>To determine the correct matches between List I and List II, we need to consider the dimensional formulas of the given physical quantities.</p>
<ol>
<li>Torque (τ) is defined as the product of force (F) and distance (d):</li>
</ol>
<p>$ \tau = F \cdot d $</p>
<p>The dimensional formula for torque is:</p>
<p>$ [\t... | mcq | jee-main-2024-online-6th-april-morning-shift | 12,865 |
bG5J7nFuAXdARcvv | physics | units-and-measurements | errors-in-measurements | A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms<sup>−1</sup>. The magnitude of its momentum is recorded as | [{"identifier": "A", "content": "17.6 kg ms<sup>-1</sup>"}, {"identifier": "B", "content": "17.565 kg ms<sup>-1</sup>"}, {"identifier": "C", "content": "17.56 kg ms<sup>-1</sup>"}, {"identifier": "D", "content": "17.57 kg ms<sup>-1</sup>"}] | ["A"] | null | Momentum, p = m $$ \times $$ v
<br/><br/>
= 3.513 $$ \times $$ 5.00 = 17.565 kg m/s
<br/><br/>
$$ \simeq $$ 17.6 kg m/s
<br/><br/>
Here number of significant digits in m is 4 and in v is 3, so, p must have minimum (which is 3) significant digit.
<br/><br/>
<b>Note :</b><br/><br/>
In this case, since we are calculating ... | mcq | aieee-2008 | 12,866 |
QJ0Wnabg6uWRis74 | physics | units-and-measurements | errors-in-measurements | The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 $$ \times $$ 10<sup>–3</sup> are | [{"identifier": "A", "content": "5, 1, 2"}, {"identifier": "B", "content": "5, 1, 5"}, {"identifier": "C", "content": "5, 5, 2"}, {"identifier": "D", "content": "4, 4, 2"}] | ["A"] | null | 23.023 has 5 significant figures.
<br><br>0.0003 has 1 significant figure.
<br><br>2.1 $$ \times $$ 10<sup>–3</sup> has 2 significant figures. | mcq | aieee-2010 | 12,867 |
1WO1uKabShEMiYbW | physics | units-and-measurements | errors-in-measurements | Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference
applied across it. If the percentage errors in the measurement of the current and the voltage difference
are 3% each, then error in the value of resistance of the wire is | [{"identifier": "A", "content": "6 %"}, {"identifier": "B", "content": "zero"}, {"identifier": "C", "content": "1 %"}, {"identifier": "D", "content": "3 %"}] | ["A"] | null | We know R = $${V \over I}$$
<br><br>$$\therefore$$ $${{\Delta R} \over R} = {{\Delta V} \over V} + {{\Delta I} \over I}$$
<br><br>Percentage error in R =
<br><br>$${{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100$$
<br><br>$$\therefore$$ $${{\Delta R} \over R} \times 1... | mcq | aieee-2012 | 12,868 |
rNK6NSjY2AtVlzUW | physics | units-and-measurements | errors-in-measurements | A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be : | [{"identifier": "A", "content": "92 $$ \\pm $$ 1.8 s"}, {"identifier": "B", "content": "92 $$ \\pm $$ 3 s"}, {"identifier": "C", "content": "92 $$ \\pm $$ 2 s"}, {"identifier": "D", "content": "92 $$ \\pm $$ 5.0 s"}] | ["C"] | null | Here t<sub>1</sub> = 90 s, t<sub>2</sub> = 91 s, t<sub>3</sub> = 95 s, t<sub>4</sub> = 92 s
<br><br>Mean(t) = $${{{t_1} + {t_2} + {t_3} + {t_4}} \over 4}$$
<br><br>= $${{90 + 91 + 95 + 92} \over 4}$$ = 92 s
<br><br>Now mean deviation
<br><br>= $${{2 + 1 + 3 + 0} \over 4}$$ = 1.5 s
<br><br>Since least count of clock is ... | mcq | jee-main-2016-offline | 12,869 |
WIwmkYyfbaRmEFOZ | physics | units-and-measurements | errors-in-measurements | The following observations were taken for determining surface tension T of water by capillary method:<br/>
diameter of capillary, D = 1.25 $$\times$$ 10<sup>-2</sup> m<br/>
rise of water, h = 1.45 $$\times$$ 10<sup>-2</sup>m<br/>
Using g = 9.80 m/s<sup>2</sup> and the simplified relation T = $${{rhg} \over 2} \times {1... | [{"identifier": "A", "content": "10 %"}, {"identifier": "B", "content": "0.15 % "}, {"identifier": "C", "content": "1.5 %"}, {"identifier": "D", "content": "2.4 %"}] | ["C"] | null | Surface tension,
<br><br>T = $${{rhg} \over 2} \times {10^3}N/m$$
<br><br>Relative error,
<br><br>$${{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}$$
<br><br>Percentage error,
<br><br>$${{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100$$
<br><br>$${{\... | mcq | jee-main-2017-offline | 12,870 |
48SnUPA0bQRSXeluiylbq | physics | units-and-measurements | errors-in-measurements | A physical quantity P is described by the relation
<br/><br/>P = a$$^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}$$ b<sup>2</sup> c<sup>3</sup> d<sup>$$-$$4</sup>
<br/><br/>If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and ... | [{"identifier": "A", "content": "8%"}, {"identifier": "B", "content": "12%"}, {"identifier": "C", "content": "32%"}, {"identifier": "D", "content": "25%"}] | ["C"] | null | Given,
<br><br>P = a$$^{{1 \over 2}}$$ b<sup>2</sup><sup></sup> c<sup>3</sup> d<sup>$$-$$4</sup>
<br><br>Relative error =
<br><br>$${{\Delta P} \over P}$$ $$ \times $$ 100 = ($${1 \over 2}$$ $$ \times $$ $${{\Delta a} \over a}$$ + 2$${{\Delta b} \over b}$$ + 3$${{\Delta c} \over c}$$ + 4$${{\Delta d} \over d}$$) $$ \... | mcq | jee-main-2017-online-9th-april-morning-slot | 12,871 |
JZRIaTnHbyjwerrd | physics | units-and-measurements | errors-in-measurements | The density of a material in the shape of a cube is determined by measuring three sides of the cube and its
mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum
error in determining the density is: | [{"identifier": "A", "content": "6%"}, {"identifier": "B", "content": "2.5%"}, {"identifier": "C", "content": "3.5%"}, {"identifier": "D", "content": "4.5%"}] | ["D"] | null | Density of a material (d) = $${M \over {{L^3}}}$$
<br><br>$$\therefore$$ Error in density,$${{\Delta d} \over d} = {{\Delta M} \over M} + 3{{\Delta L} \over L}$$
<br><br>$${{\Delta d} \over d} \times 100 = {{\Delta M} \over M} \times 100 + 3{{\Delta L} \over L} \times 100$$
<br><br>$$ \Rightarrow {{\Delta d} \over d} \... | mcq | jee-main-2018-offline | 12,872 |
uNKq0YRE3wVHutE4Kphwd | physics | units-and-measurements | errors-in-measurements | The relative error in the determination of the surface area of sphere is $$\alpha $$. Then the relative error in the determination of its volume is : | [{"identifier": "A", "content": "$${3 \\over 2}\\alpha $$"}, {"identifier": "B", "content": "$${2 \\over 3}\\alpha $$"}, {"identifier": "C", "content": "$${5 \\over 2}\\alpha $$"}, {"identifier": "D", "content": "$$\\alpha $$"}] | ["A"] | null | Relative error in the surface are of the sphere,
<br><br>$${{\Delta S} \over S}$$ = 2 $$ \times $$ $${{\Delta r} \over r}$$ = $$ \propto $$ (given)
<br><br>Relative error in volume,
<br><br>$${{\Delta V} \over V}$$ = 3 $$ \times $$ $${{\Delta r} \over r}$$
<br><br>= 3 $$ \times $$ $${1 \over 2}$$ $$ \times $$ $${{\Del... | mcq | jee-main-2018-online-15th-april-morning-slot | 12,873 |
stgRK8lBApemvKm4g3YKp | physics | units-and-measurements | errors-in-measurements | The relative uncertainly in the period of a satellite orbiting around the earth is 10<sup>-2</sup>. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is : | [{"identifier": "A", "content": "10<sup>$$-$$2</sup>"}, {"identifier": "B", "content": "2 $$ \\times $$ 10<sup>$$-$$2</sup>"}, {"identifier": "C", "content": "3 $$ \\times $$ 10<sup>$$-$$2</sup>"}, {"identifier": "D", "content": "6 $$ \\times $$ 10<sup>$$-$$2</sup>"}] | ["B"] | null | From kepler's law,
<br><br>T = 2$$\pi $$ $$\sqrt {{{{r^3}} \over {GM}}} $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$T<sup>2</sup> = $${{4{\pi ^2}} \over {GM}}{r^3}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ M = $${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$$
<br><br>$$\therefore\,\,\,$$ $${{\Delta M} \over M}$$ = 2 $${{\... | mcq | jee-main-2018-online-16th-april-morning-slot | 12,875 |
59uIoYKOVAjaPOyoFRGtx | physics | units-and-measurements | errors-in-measurements | In a simple pendulum experiment for
determination of acceleration due to gravity (g),
time taken for 20 oscillations is measured by
using a watch of 1 second least count. The
mean value of time taken comes out to be
30 s. The length of pendulum is measured by
using a meter scale of least count 1 mm and the
value obtain... | [{"identifier": "A", "content": "0.2%"}, {"identifier": "B", "content": "3.5%"}, {"identifier": "C", "content": "0.7%"}, {"identifier": "D", "content": "6.8%"}] | ["D"] | null | Time period of a pendulum (T) = $$2\pi \sqrt {{l \over g}} $$
<br><br>$$ \Rightarrow $$ T<sup>2</sup> = $$4{\pi ^2}{l \over g}$$
<br><br>$$ \Rightarrow $$ $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
<br><br>Fractional change
<br><br>$$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 - \... | mcq | jee-main-2019-online-8th-april-evening-slot | 12,876 |
O96L41Np317XvIrPuQQIp | physics | units-and-measurements | errors-in-measurements | In the density measurement of a cube, the mass and
edge length are measured as (10.00 ± 0.10) kg and
(0.10 ± 0.01) m, respectively. The error in the
measurement of density is : | [{"identifier": "A", "content": "0.01 kg/m<sup>3</sup>"}, {"identifier": "B", "content": "0.10 kg/m<sup>3</sup>"}, {"identifier": "C", "content": "0.31 kg/m<sup>3</sup>"}, {"identifier": "D", "content": "0.07 kg/m<sup>3</sup>"}] | ["C"] | null | Mass (m) = (10.00 ± 0.10) kg
<br><br>Edge length ($$l$$) = (0.10 ± 0.01) m
<br><br>Volume of the cube (V) = $${l^3}$$
<br><br>Density, $$\rho $$ = $${m \over V}$$
<br><br>$${{d\rho } \over \rho } = {{dm} \over m} + {{dV} \over V}$$
<br><br>$$ \Rightarrow $$ $${{d\rho } \over \rho } = {{dm} \over m} + 3{{dl} \over l}$$
... | mcq | jee-main-2019-online-9th-april-morning-slot | 12,878 |
l3NOPR5ufPBk0BNChP3fB | physics | units-and-measurements | errors-in-measurements | The diameter and height of a cylinder are measured by a meter scale to be 12.6 $$ \pm $$ 0.1 cm and 34.2 $$ \pm $$ 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ? | [{"identifier": "A", "content": "4264.4 $$ \\pm $$ 81.0 cm<sup>3</sup>"}, {"identifier": "B", "content": "4264 $$ \\pm $$ 81 cm<sup>3</sup>\n"}, {"identifier": "C", "content": "4300 $$ \\pm $$ 80 cm<sup>3</sup>\n"}, {"identifier": "D", "content": "4260 $$ \\pm $$ 80 cm<sup>3</sup>"}] | ["D"] | null | Volume of cylinder(V) = $$\pi $$r<sup>2</sup>h
<br><br>= $$\pi {{{d^2}} \over 4}h$$
<br><br>= $$3.14 \times {{{{\left( {12.6} \right)}^2}} \over 4} \times 34.2$$
<br><br> = 4260
<br><br>$${{\Delta V} \over V} = 2{{\Delta d} \over d} + {{\Delta h} \over h} = 2\left( {{{0.1} \over {12.6}}} \right) + {{0.1} \over {34.2}... | mcq | jee-main-2019-online-10th-january-evening-slot | 12,879 |
fLULH050rO7T9paNTwB92 | physics | units-and-measurements | errors-in-measurements | A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is : | [{"identifier": "A", "content": "2.0 %"}, {"identifier": "B", "content": "2.5 %"}, {"identifier": "C", "content": "1.0 %"}, {"identifier": "D", "content": "0.5 %"}] | ["C"] | null | We know,
<br><br>$$R = {{\rho l} \over A}$$
<br><br>and Volume (V) = A$$l$$
<br><br>$$ \Rightarrow $$ A $$=$$ $${V \over l}$$
<br><br>$$ \therefore $$ $$R = {{\rho {l^2}} \over v}$$
<br><br>$$ \therefore $$ $${{\Delta R} \over R} = 2{{\Delta l} \over l}$$
<br><br>$$=$$... | mcq | jee-main-2019-online-9th-january-morning-slot | 12,880 |
eob1dkaXXRsAwz9PFzjgy2xukfruwy16 | physics | units-and-measurements | errors-in-measurements | The density of a solid metal sphere is determined by measuring its mass and its diameter. The
maximum error in the density of the sphere is $$\left( {{x \over {100}}} \right)$$ %. If the relative errors in measuring the mass
and the diameter are 6.0% and 1.5% respectively, the value of x is_______. | [] | null | 1050 | $$\rho $$ = $${M \over V}$$ = $${M \over {{4 \over 3}\pi {{\left( {{D \over 2}} \right)}^3}}}$$
<br><br>$$ \Rightarrow $$ $$\rho $$ = $${6 \over \pi }M{D^{ - 3}}$$
<br><br>For maximum error
<br><br>$${{d\rho } \over \rho } \times 100 = {{dM} \over M} \times 100 + 3{{dD} \over D} \times 100$$
<br><br>= 6 + 3 $$ \times $... | integer | jee-main-2020-online-6th-september-morning-slot | 12,883 |
aDfRRaQHiCzjB0ysBY1kmhphl8i | physics | units-and-measurements | errors-in-measurements | The resistance R = $${V \over I}$$, where V = (50 $$\pm$$ 2)V and I = (20 $$\pm$$ 0.2)A. The percentage error in R is 'x'%. The value of 'x' to the nearest integer is _________. | [] | null | 5 | $$R = {V \over I}$$<br><br>$${{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100$$<br><br>% error in $$R = {2 \over {50}} \times 100 + {{0.2} \over {20}} \times 100$$<br><br>% error in R = 4 + 1<br><br>$$ \therefore $$ % error in R = 5% | integer | jee-main-2021-online-16th-march-morning-shift | 12,884 |
6KPZcoU8UJSSsYtZXq1kmipb8hp | physics | units-and-measurements | errors-in-measurements | In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of ... | [{"identifier": "A", "content": "0.14%"}, {"identifier": "B", "content": "9%"}, {"identifier": "C", "content": "1.4%"}, {"identifier": "D", "content": "0.9%"}] | ["C"] | null | $${{\Delta Y} \over Y} = \left( {{{\Delta m} \over m}} \right) + \left( {{{\Delta g} \over g}} \right) + \left( {{{\Delta A} \over A}} \right) + \left( {{{\Delta l} \over l}} \right) + \left( {{{\Delta L} \over L}} \right)$$<br><br>$$ = \left( {{{1g} \over {1kg}}} \right) + 0 + 2\left( {{{\Delta r} \over r}} \right) + ... | mcq | jee-main-2021-online-16th-march-evening-shift | 12,885 |
cFgnWxkJueEBysJRpp1kmlwtmsr | physics | units-and-measurements | errors-in-measurements | The radius of a sphere is measured to be (7.50 $$\pm$$ 0.85) cm. Suppose the percentage error in its volume is x. <br/><br/>The value of x, to the nearest x, is __________. | [] | null | 34 | Given, radius of sphere = (7.50 $$\pm$$ 0.85) cm<br><br>$$ \therefore $$ r = 7.50 and dr = 0.85<br><br>We know, volume of a sphere $$v = {4 \over 3}\pi {r^3}$$<br><br>taking log both sides, we get<br><br>$$\ln v = \ln {{4\pi } \over 3} + 3\ln r$$<br><br>Differentiating both sides,<br><br>$${{dv} \over v} = 0 + 3{{dr} \... | integer | jee-main-2021-online-18th-march-evening-shift | 12,886 |
1krsx2xz4 | physics | units-and-measurements | errors-in-measurements | Three students S<sub>1</sub>, S<sub>2</sub> and S<sub>3</sub> perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.<br/><br/><table>
<thea... | [] | null | 1 | $$T = {t \over n} = 2\pi \sqrt {{l \over g}} $$<br><br>$$ \Rightarrow g = {{4{\pi ^2}l} \over {{T^2}}}$$<br><br>$$ \Rightarrow {{\Delta g} \over g} \times 100 = {{\Delta l} \over l} \times 100 + 2{{\Delta T} \over T} \times 100$$<br><br>$$ = \left( {{{\Delta l} \over l} + {{2\Delta T} \over {T}}} \right)100\% $$<br><br... | integer | jee-main-2021-online-22th-july-evening-shift | 12,887 |
1ks18vtz1 | physics | units-and-measurements | errors-in-measurements | A physical quantity 'y' is represented by the formula $$y = {m^2}{r^{ - 4}}{g^x}{l^{ - {3 \over 2}}}$$<br/><br/>If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p. | [{"identifier": "A", "content": "5 and $$\\pm$$2"}, {"identifier": "B", "content": "4 and $$\\pm$$3"}, {"identifier": "C", "content": "$${{16} \\over 3}$$ and $$ \\pm {3 \\over 2}$$"}, {"identifier": "D", "content": "8 and $$\\pm$$ 2"}] | ["C"] | null | $${{\Delta y} \over y} = {{2\Delta m} \over m} + {{4\Delta r} \over r} + {{x\Delta g} \over g} + {3 \over 2}{{\Delta l} \over l}$$<br><br>$$18 = 2(1) + 4(0.5) + xp + {3 \over 2}(4)$$<br><br>$$ \Rightarrow $$ 8 = xp<br><br>By checking from options.<br><br>$$x = {{16} \over 3},p = \pm {3 \over 2}$$ | mcq | jee-main-2021-online-27th-july-evening-shift | 12,888 |
1ktbrll2z | physics | units-and-measurements | errors-in-measurements | If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is : | [{"identifier": "A", "content": "86.4 s"}, {"identifier": "B", "content": "4.32 s"}, {"identifier": "C", "content": "43.2 s"}, {"identifier": "D", "content": "8.64 s"}] | ["C"] | null | $$T = 2\pi \sqrt {{l \over g}} $$<br><br>$${{\Delta T} \over T} = {1 \over 2}{{\Delta l} \over l}$$<br><br>$$\Delta T = {1 \over 2} \times {{0.1} \over {100}} \times 24 \times 3600$$<br><br>$$\Delta T = 43.2$$ | mcq | jee-main-2021-online-26th-august-evening-shift | 12,889 |
1ktbv72jc | physics | units-and-measurements | errors-in-measurements | The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..............% | [] | null | 14 | $$T = 2\pi \sqrt {{l \over g}} \Rightarrow l = {{{T^2}g} \over {4{\pi ^2}}}$$<br><br>$$E = mgl{{{\theta ^2}} \over 2} = m{g^2}{{{T^2}{\theta ^2}} \over {8{\pi ^2}}}$$<br><br>$${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} \right)$$<br><br>$$ = (4 + 3) = 14\% $$ | integer | jee-main-2021-online-26th-august-evening-shift | 12,890 |
1ktmo4lfv | physics | units-and-measurements | errors-in-measurements | A student determined Young's Modulus of elasticity using the formula $$Y = {{Mg{L^3}} \over {4b{d^3}\delta }}$$. The value of g is taken to be 9.8 m/s<sup>2</sup>, without any significant error, his observation are as following.<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%... | [{"identifier": "A", "content": "0.0083"}, {"identifier": "B", "content": "0.0155"}, {"identifier": "C", "content": "0.155"}, {"identifier": "D", "content": "0.083"}] | ["B"] | null | $$y = {{Mg{L^3}} \over {4b{d^3}\delta }}$$<br><br>$${{\Delta y} \over y} = {{\Delta M} \over M} + {{3\Delta L} \over L} + {{\Delta b} \over b} + {{3\Delta d} \over d} + {{\Delta \delta } \over \delta }$$<br><br>$${{\Delta y} \over y} = {{{{10}^{ - 3}}} \over 2} + {{3 \times {{10}^{ - 3}}} \over 1} + {{{{10}^{ - 2}}} \o... | mcq | jee-main-2021-online-1st-september-evening-shift | 12,891 |
1l55ls2f1 | physics | units-and-measurements | errors-in-measurements | <p>A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is $${x \over {121}}\% $$. The value of x is ____________.</p> | [] | null | 150 | <p>$${I_{mean}} = {{1.22 + 1.23 + 1.19 + 1.20} \over 4} = 1.21$$</p>
<p>$$\Delta {I_{mean}} = {{0.01 + 0.02 + 0.02 + 0.01} \over 4} = 0.015$$</p>
<p>So % $$I = {{\Delta {I_{mean}}} \over {{I_{mean}}}} \times 100 = {{0.015} \over {1.21}} \times 100 = {{150} \over {121}}\% $$</p>
<p>$$x = 150$$</p> | integer | jee-main-2022-online-28th-june-evening-shift | 12,892 |
1l57ph8j3 | physics | units-and-measurements | errors-in-measurements | <p>A silver wire has a mass (0.6 $$\pm$$ 0.006) g, radius (0.5 $$\pm$$ 0.005) mm and length (4 $$\pm$$ 0.04) cm. The maximum percentage error in the measurement of its density will be :</p> | [{"identifier": "A", "content": "4%"}, {"identifier": "B", "content": "3%"}, {"identifier": "C", "content": "6%"}, {"identifier": "D", "content": "7%"}] | ["A"] | null | <p>$$\rho = {m \over V} = {m \over {\pi {r^2} \times l}}$$</p>
<p>$$\therefore$$ % error in $$\rho = \left( {{{0.006} \over {0.6}} + 2 \times {{0.005} \over {0.5}} + {{0.04} \over 4}} \right) \times 100$$</p>
<p>$$ = 4\% $$</p> | mcq | jee-main-2022-online-27th-june-morning-shift | 12,893 |
1l59ppqqu | physics | units-and-measurements | errors-in-measurements | <p>For $$z = {a^2}{x^3}{y^{{1 \over 2}}}$$, where 'a' is a constant. If percentage error in measurement of 'x' and 'y' are 4% and 12% respectively, then the percentage error for 'z' will be _______________%.</p> | [] | null | 18 | <p>% error in $$z = 3 \times 4 + {1 \over 2} \times 12$$</p>
<p>$$ = 12 + 6 = 18\% $$</p> | integer | jee-main-2022-online-25th-june-evening-shift | 12,894 |
1l5ak50l1 | physics | units-and-measurements | errors-in-measurements | <p>If $$Z = {{{A^2}{B^3}} \over {{C^4}}}$$, then the relative error in Z will be :</p> | [{"identifier": "A", "content": "$${{\\Delta A} \\over A} + {{\\Delta B} \\over B} + {{\\Delta C} \\over C}$$"}, {"identifier": "B", "content": "$${{2\\Delta A} \\over A} + {{3\\Delta B} \\over B} - {{4\\Delta C} \\over C}$$"}, {"identifier": "C", "content": "$${{2\\Delta A} \\over A} + {{3\\Delta B} \\over B} + {{4\\D... | ["C"] | null | <p>$$Z = {{{A^2}{B^3}} \over {{C^4}}}$$</p>
<p>$$\therefore$$ $${{\Delta Z} \over Z} = {{2\Delta A} \over A} + 3 \times {{\Delta B} \over B} + {{4\Delta C} \over C}$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift | 12,895 |
1l6jepz9i | physics | units-and-measurements | errors-in-measurements | <p>A torque meter is calibrated to reference standards of mass, length and time each with $$5 \%$$ accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :</p> | [{"identifier": "A", "content": "15%"}, {"identifier": "B", "content": "25%"}, {"identifier": "C", "content": "75%"}, {"identifier": "D", "content": "5%"}] | ["B"] | null | <p>We know that, torque applied on a rotating body,
<br/><br/>$$
\begin{aligned}
\tau &=\text { Force } \times \text { Perpendicular distance } \\\\
\Rightarrow \quad[\tau] &=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \Rightarrow[\tau]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
$$</p>
<p>$$ \Rightarrow {... | mcq | jee-main-2022-online-27th-july-morning-shift | 12,897 |
1l6nu8u7x | physics | units-and-measurements | errors-in-measurements | <p>In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $$0.5 \mathrm{~s}$$ is measured from time of 100 oscillation with a watch of $$1 \mathrm{~s}$$ resolution. If measured value of length is $$10 \mathrm{~cm}$$ known to $$1 \mathrm{~mm}$$ accuracy, The accuracy in the determ... | [] | null | 5 | <p>$$T = 2\pi \sqrt {{l \over g}} $$</p>
<p>$${{dg} \over g} \times 100 = {{2dT} \over T} \times 100 + {{dl} \over l} \times 100$$</p>
<p>$$ = 2 \times {1 \over {50}} \times 100 + {1 \over {100}} \times 100 = 5\% $$</p> | integer | jee-main-2022-online-28th-july-evening-shift | 12,898 |
1lgq2s5sc | physics | units-and-measurements | errors-in-measurements | <p>A body of mass $$(5 \pm 0.5) ~\mathrm{kg}$$ is moving with a velocity of $$(20 \pm 0.4) ~\mathrm{m} / \mathrm{s}$$. Its kinetic energy will be</p> | [{"identifier": "A", "content": "$$(1000 \\pm 140) ~\\mathrm{J}$$"}, {"identifier": "B", "content": "$$(500 \\pm 0.14) ~\\mathrm{J}$$"}, {"identifier": "C", "content": "$$(1000 \\pm 0.14) ~\\mathrm{J}$$"}, {"identifier": "D", "content": "$$(500 \\pm 140) ~\\mathrm{J}$$"}] | ["A"] | null | To find the kinetic energy of the body, we can use the formula:
<br/><br/>
$$KE = \frac{1}{2}mv^2$$
<br/><br/>
Given the mass $$m = (5 \pm 0.5) \,\text{kg}$$ and the velocity $$v = (20 \pm 0.4) \,\text{m/s}$$, we can find the kinetic energy and its uncertainty by applying the rules for the propagation of errors in mult... | mcq | jee-main-2023-online-13th-april-morning-shift | 12,899 |
1lgxwqesk | physics | units-and-measurements | errors-in-measurements | <p>A physical quantity P is given as</p>
<p>$$P = {{{a^2}{b^3}} \over {c\sqrt d }}$$<?p>
<p>The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be</p></p> | [{"identifier": "A", "content": "12%"}, {"identifier": "B", "content": "13%"}, {"identifier": "C", "content": "16%"}, {"identifier": "D", "content": "14%"}] | ["B"] | null | <p>The percentage error in a quantity that is a product or quotient of other quantities is given by the sum of the percentage errors in those quantities, each multiplied by the power to which it is raised in the expression for the quantity.</p>
<p>Given the physical quantity P as </p>
<p>$$P = \frac{a^2b^3}{c\sqrt{d}}$... | mcq | jee-main-2023-online-10th-april-morning-shift | 12,900 |
1lh01bfcp | physics | units-and-measurements | errors-in-measurements | <p>A cylindrical wire of mass $$(0.4 \pm 0.01) \mathrm{g}$$ has length $$(8 \pm 0.04) \mathrm{cm}$$ and radius $$(6 \pm 0.03) \mathrm{mm}$$. The maximum error in its density will be:</p> | [{"identifier": "A", "content": "1%"}, {"identifier": "B", "content": "5%"}, {"identifier": "C", "content": "4%"}, {"identifier": "D", "content": "3.5%"}] | ["C"] | null | <p>The density of a cylindrical wire is given by the formula:</p>
<p>$$\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$$</p>
<p>where $m$ is the mass, $r$ is the radius, and $l$ is the length. </p>
<p>The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on. For the density... | mcq | jee-main-2023-online-8th-april-morning-shift | 12,901 |
1lh25ck1e | physics | units-and-measurements | errors-in-measurements | <p>Two resistances are given as $$\mathrm{R}_{1}=(10 \pm 0.5) \Omega$$ and $$\mathrm{R}_{2}=(15 \pm 0.5) \Omega$$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -</p> | [{"identifier": "A", "content": "2.33"}, {"identifier": "B", "content": "5.33"}, {"identifier": "C", "content": "4.33"}, {"identifier": "D", "content": "6.33"}] | ["C"] | null | <p>In the problem, we are given two resistances, $R_1$ and $R_2$, each with a certain measurement error, $\Delta R_1$ and $\Delta R_2$. These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.</p>
<p>The formula for the equivalent resis... | mcq | jee-main-2023-online-6th-april-morning-shift | 12,902 |
lsbl30sw | physics | units-and-measurements | errors-in-measurements | The radius $(\mathrm{r})$, length $(l)$ and resistance $(\mathrm{R})$ of a metal wire was measured in the laboratory as<br/><br/>
$$
\begin{aligned}
& \mathrm{r}=(0.35 \pm 0.05) ~\mathrm{cm} \\\\
& \mathrm{R}=(100 \pm 10) ~\mathrm{ohm} \\\\
& l=(15 \pm 0.2)~ \mathrm{cm}
\end{aligned}
$$
<br/><br/>
The perce... | [{"identifier": "A", "content": "$37.3 \\%$"}, {"identifier": "B", "content": "$25.6 \\%$"}, {"identifier": "C", "content": "$35.6 \\%$"}, {"identifier": "D", "content": "$39.9 \\%$"}] | ["D"] | null | <p>To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity. The resistivity $$ \rho $$ of a wire is given by:</p>
<p>$\rho = \frac{RA}{l}$</p>
<p>where:</p>
<ul>
<li>$$ R $$ is the resistance</li><br>
<li>$$ A $$ is the cross-sectional a... | mcq | jee-main-2024-online-1st-february-morning-shift | 12,904 |
jaoe38c1lsd69lms | physics | units-and-measurements | errors-in-measurements | <p>The measured value of the length of a simple pendulum is $$20 \mathrm{~cm}$$ with $$2 \mathrm{~mm}$$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $$\mathrm{N} \%$$. The value o... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\
& \mathrm{g}=\frac{4 \pi^2 \ell}{\mathrm{T}^2} \\
& \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \\
& =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\
& =\frac{1.2}{20}
\end{aligned}$$</p>
<p>P... | mcq | jee-main-2024-online-31st-january-evening-shift | 12,905 |
jaoe38c1lse6k1yn | physics | units-and-measurements | errors-in-measurements | <p>If the percentage errors in measuring the length and the diameter of a wire are $$0.1 \%$$ each. The percentage error in measuring its resistance will be:</p> | [{"identifier": "A", "content": "0.144%"}, {"identifier": "B", "content": "0.2%"}, {"identifier": "C", "content": "0.1%"}, {"identifier": "D", "content": "0.3%"}] | ["D"] | null | <p>$$\begin{aligned}
& \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{\mathrm{d}^2}{4}} \\
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}} \\
& \frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \% \text { and } \frac{\Delta \mathrm{d}}{\mathrm{d}}=0.1 \% \\
& \frac... | mcq | jee-main-2024-online-31st-january-morning-shift | 12,906 |
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