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|---|---|---|---|---|---|---|---|---|---|---|---|
1ktdz96vv | physics | circular-motion | uniform-circular-motion | A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second?<br/><br/>Given : 1 ly = 9.46 $$\times$$ 10<sup>15</sup> m<br/><br/>1 AU = 1.5 $$\times$$ 10<sup>11</sup> m | [{"identifier": "A", "content": "4.1 $$\\times$$ 10<sup>8</sup> s"}, {"identifier": "B", "content": "4.5 $$\\times$$ 10<sup>10</sup> s"}, {"identifier": "C", "content": "3.5 $$\\times$$ 10<sup>6</sup> s"}, {"identifier": "D", "content": "7.2 $$\\times$$ 10<sup>8</sup> s"}] | ["B"] | null | $$R = {l \over \theta }$$<br><br>Time $$ = {{4 \times 2\pi R} \over v} = {{4 \times 2\pi } \over v}\left( {{l \over \theta }} \right)$$<br><br>put l = 4.4 $$\times$$ 9.46 $$\times$$ 10<sup>15</sup><br><br>v = 8 $$\times$$ 1.5 $$\times$$ 10<sup>11</sup><br><br>$$\theta = {4 \over {3600}} \times {\pi \over {180}}$$ rad... | mcq | jee-main-2021-online-27th-august-morning-shift | 9,536 |
1l58beb40 | physics | circular-motion | uniform-circular-motion | <p>A ball is released from rest from point P of a smooth semi-spherical vessel as shown in figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is $$\alpha$$ with respect to point P. Which of the following graphs represent the correct relation betw... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5bp6iwl/8caae413-17b9-45fa-a71c-27b010be4ac2/cbaecf40-fe51-11ec-b169-b5046c590266/file-1l5bp6iwm.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l5bp6iwl/8caae413-17b9-45fa-a71c-27b010be4ac2/cba... | ["C"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5jhr5j1/302ab3e9-b1f4-4544-b636-4c3b0160e5f3/058126d0-029b-11ed-a9b8-43edceee002f/file-1l5jhr5j2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5jhr5j1/302ab3e9-b1f4-4544-b636-4c3b0160e5f3/058126d0-029b-11ed-a9b8-43edceee002... | mcq | jee-main-2022-online-26th-june-morning-shift | 9,537 |
1l59lxvo1 | physics | circular-motion | uniform-circular-motion | <p>A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $$\omega$$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $$\mu$$. The... | [{"identifier": "A", "content": "$$R \\le {{\\mu g} \\over {2{\\omega ^2}}}$$"}, {"identifier": "B", "content": "$$R \\le {{\\mu g} \\over {{\\omega ^2}}}$$"}, {"identifier": "C", "content": "$$R \\ge {{\\mu g} \\over {2{\\omega ^2}}}$$"}, {"identifier": "D", "content": "$$R \\ge {{\\mu g} \\over {{\\omega ^2}}}$$"}] | ["B"] | null | <p>The force that prevents the beaker from sliding is the static friction force. For an object in uniform circular motion, the net force acting on the object (which is the friction force in this case) is equal to the centripetal force.</p>
<p>The static friction force is given by the normal force (which is equal to the... | mcq | jee-main-2022-online-25th-june-evening-shift | 9,538 |
1l59m92bx | physics | circular-motion | uniform-circular-motion | <p>For a particle in uniform circular motion, the acceleration $$\overrightarrow a $$ at any point P(R, $$\theta$$) on the circular path of radius R is (when $$\theta$$ is measured from the positive x-axis and v is uniform speed) :</p> | [{"identifier": "A", "content": "$$ - {{{v^2}} \\over R}\\sin \\theta \\widehat i + {{{v^2}} \\over R}\\cos \\theta \\widehat j$$"}, {"identifier": "B", "content": "$$ - {{{v^2}} \\over R}\\cos \\theta \\widehat i + {{{v^2}} \\over R}\\sin \\theta \\widehat j$$"}, {"identifier": "C", "content": "$$ - {{{v^2}} \\over R}... | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5kpo2ny/38be5e44-dc63-4c88-be0c-45cfdf5a0089/c18c12d0-0346-11ed-8429-27c3de0e24ff/file-1l5kpo2nz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5kpo2ny/38be5e44-dc63-4c88-be0c-45cfdf5a0089/c18c12d0-0346-11ed-8429-27c3de0e24ff... | mcq | jee-main-2022-online-25th-june-evening-shift | 9,539 |
1l59pr8gl | physics | circular-motion | uniform-circular-motion | <p>A curved in a level road has a radius 75 m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be ___________ m/s.</p> | [] | null | 24 | <p>$$\because$$ $$v = \sqrt {\mu gr} $$</p>
<p>$$ \Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{{{r_1}} \over {{r_2}}}} $$</p>
<p>$$ \Rightarrow {{30} \over {{v_2}}} = \sqrt {{{75} \over {48}}} = \sqrt {{{25} \over {16}}} = {5 \over 4}$$</p>
<p>$$ \Rightarrow {V_2} = 24$$ m/s</p> | integer | jee-main-2022-online-25th-june-evening-shift | 9,540 |
1l5bbi2oh | physics | circular-motion | uniform-circular-motion | <p>A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is</p> | [{"identifier": "A", "content": "the same throughout the motion."}, {"identifier": "B", "content": "minimum at the highest position of the circular path."}, {"identifier": "C", "content": "minimum at the lowest position of the circular path."}, {"identifier": "D", "content": "minimum when the rope is in the horizontal ... | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5l6hdcv/7e2054d6-02e0-4296-bf08-104db7f196ee/8256b6f0-0388-11ed-bfd1-873560f9d960/file-1l5l6hdcw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5l6hdcv/7e2054d6-02e0-4296-bf08-104db7f196ee/8256b6f0-0388-11ed-bfd1-873560f9d960... | mcq | jee-main-2022-online-24th-june-evening-shift | 9,541 |
1l5bbzc2e | physics | circular-motion | uniform-circular-motion | <p>A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be :</p> | [{"identifier": "A", "content": "7.5 rad"}, {"identifier": "B", "content": "15 rad"}, {"identifier": "C", "content": "20 rad"}, {"identifier": "D", "content": "30 rad"}] | ["B"] | null | <p>$${\theta _1} = {1 \over 2}\alpha (2 \times 1 - 1) = 5$$ rad</p>
<p>$$\Rightarrow$$ $$\alpha$$ = 10 rad/sec<sup>2</sup></p>
<p>So $${\theta _2} = {1 \over 2} \times \alpha (2 \times 2 - 1) = 15$$ rad</p> | mcq | jee-main-2022-online-24th-june-evening-shift | 9,542 |
1l5w3akjf | physics | circular-motion | uniform-circular-motion | <p>A person starts his journey from centre 'O' of the park and comes back to the same position following path OPQO as shown in the figure. The radius of path taken by the person is 200 m and he takes 3 min 58 sec to complete his journey. The average speed of the person is _____________ ms<sup>$$-$$1</sup>. (take $$\pi$... | [] | null | 3 | $$
\begin{aligned}
& V_{a v}=\frac{S_{\text {total }}}{T_{\text {total }}}=\frac{200+200+\frac{2 \pi \times 200}{4}}{238} \\\\
&=\frac{400+100 \pi}{238}=\frac{714}{238}=3 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$ | integer | jee-main-2022-online-30th-june-morning-shift | 9,544 |
1l6dxvr5w | physics | circular-motion | uniform-circular-motion | <p>A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is $$60 \mathrm{~m}$$, then the magnitude of displacement would be :</p>
<p>(Given $$\left.\cos 135^{\circ}=-0.7\right)$$</p>
<p><img src="data:image/png;base64,UklGRvQRAABXRUJQVlA4IOgRAADwJgGdASoAA9QCP4HA2mW2MC0nIdSpC... | [{"identifier": "A", "content": "42 m"}, {"identifier": "B", "content": "47 m"}, {"identifier": "C", "content": "19 m"}, {"identifier": "D", "content": "40 m"}] | ["B"] | null | <p>Distance travelled = 60 m</p>
<p>$$\Rightarrow$$ Angle covered = 135$$^\circ$$</p>
<p>Displacement $$ = 2R\sin \left( {{{135^\circ } \over 2}} \right)$$</p>
<p>$$ = 2\left( {{{60} \over {135}} \times {{180} \over \pi }} \right){\left[ {{{1 - \cos (135^\circ )} \over 2}} \right]^{1/2}}$$</p>
<p>$$ = 2\left( {{{80} \o... | mcq | jee-main-2022-online-25th-july-morning-shift | 9,545 |
1l6p43wr0 | physics | circular-motion | uniform-circular-motion | <p>A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass m moves against the wall with a speed v. Which of the following curve represents the correct relation between the normal reaction on the block by the wall (N) and speed of the block (v) ?</p>
<p><img src="data:image/png;base64,Uk... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6sadqp2/47ae5ed7-0c9d-4008-994e-63584ef5c618/75d91770-1b3d-11ed-b433-d3d77037ad84/file-1l6sadqp3.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l6sadqp2/47ae5ed7-0c9d-4008-994e-63584ef5c618/75d... | ["A"] | null | <p>Block is moving in the circular path, that means there should be a force which is keeping the block in contact with the wall here that force is normal force, which is towards the center of the circular path. And we know this force is equal to ${{m{v^2}} \over r}$.</p>
<p>$$ \therefore $$ $$N = {{m{v^2}} \over r}$$</... | mcq | jee-main-2022-online-29th-july-morning-shift | 9,546 |
ldo7daud | physics | circular-motion | uniform-circular-motion | A body is moving with constant speed, in a circle of radius $10 \mathrm{~m}$. The body completes one revolution in $4 \mathrm{~s}$. At the end of 3rd second, the displacement of body (in $\mathrm{m}$ ) from its starting point is : | [{"identifier": "A", "content": "$15 \\pi$"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "$10 \\sqrt{2}$"}, {"identifier": "D", "content": "$5 \\pi$"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{\omega}=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{4}=\frac{\pi}{2} \mathrm{rad} / \mathrm{s} \\\\
& \theta=\mathrm{\omega t} \\\\
& \theta=\frac{\pi}{2} \times 3 \\\\
& \theta=\frac{3 \pi}{2} \mathrm{rad}
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fl... | mcq | jee-main-2023-online-31st-january-evening-shift | 9,548 |
ldqwd3b1 | physics | circular-motion | uniform-circular-motion | A stone tied to $180 \mathrm{~cm}$ long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is $\frac{1936}{x} ms^{-2}$. The value of $x$ ________. (Take $\pi=\frac{22}{7}$ ) | [] | null | 125 | Acceleration of stone $a=\frac{v^{2}}{r}=\omega^{2} R$
<br/><br/>$$
\begin{aligned}
& a=\left(\frac{28 \times 2}{60} \times \frac{22}{7}\right)^{2} \times 1.8 \\\\
& =\frac{1936}{125}
\end{aligned}
$$
<br/><br/>So, $x=125$ | integer | jee-main-2023-online-30th-january-evening-shift | 9,549 |
1ldsal8zn | physics | circular-motion | uniform-circular-motion | <p>An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at $$x=+2~\mathrm{m}$$, its velocity is $$\mathrm{ - 4\widehat j}$$ m/s. The object's velocity (v) and acceleration (a) at $$x=-2~\mathrm{m}$$ will be</p> | [{"identifier": "A", "content": "$$v=4\\mathrm{\\widehat i~m/s},a=8\\mathrm{\\widehat j~m/s^2}$$"}, {"identifier": "B", "content": "$$v=4\\mathrm{\\widehat j~m/s},a=8\\mathrm{\\widehat i~m/s^2}$$"}, {"identifier": "C", "content": "$$v=-4\\mathrm{\\widehat i~m/s},a=-8\\mathrm{\\widehat j~m/s^2}$$"}, {"identifier": "D", ... | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1let35sjj/457e1035-e46d-40a0-b724-d3dc666cfc52/bc939de0-ba0f-11ed-bf62-29125e0e0d77/file-1let35sjk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1let35sjj/457e1035-e46d-40a0-b724-d3dc666cfc52/bc939de0-ba0f-11ed-bf62-29125e0e0d77... | mcq | jee-main-2023-online-29th-january-evening-shift | 9,550 |
1ldsp8hcq | physics | circular-motion | uniform-circular-motion | <p>A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [take g = 10 ms$$^{-2}$$]</p> | [{"identifier": "A", "content": "3.4 ms$$^{-1}$$"}, {"identifier": "B", "content": "13 ms$$^{-1}$$"}, {"identifier": "C", "content": "22.4 ms$$^{-1}$$"}, {"identifier": "D", "content": "17 ms$$^{-1}$$"}] | ["B"] | null | $f_{s}=\frac{m v^{2}}{r}$
<br/><br/>
For maximum speed in safe turning,
<br/><br/>
$\mathrm{f}_{\mathrm{s}}=\mathrm{f}_{\mathrm{s}} \max =\mu \mathrm{mg}$
<br/><br/>
$\mathrm{v}_{\max }$ (for safe turning) $=\sqrt{\mu \mathrm{rg}}$
<br/><br/>
$=\sqrt{0.34 \times 50 \times 10} \approx 13 \mathrm{~m} / \mathrm{s}$ | mcq | jee-main-2023-online-29th-january-morning-shift | 9,551 |
1ldui3641 | physics | circular-motion | uniform-circular-motion | <p>A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g = 10 m/s$$^2$$)</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{2}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi}{6}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{4}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi}{3}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5jfhmh/21063d45-0704-4c91-a742-33934be23a35/8cb70e90-ad1c-11ed-8bc1-d3bd0941e5b5/file-1le5jfhmi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5jfhmh/21063d45-0704-4c91-a742-33934be23a35/8cb70e90-ad1c-11ed-8bc1-d3bd0941e5b5... | mcq | jee-main-2023-online-25th-january-morning-shift | 9,552 |
1ldwrt6cw | physics | circular-motion | uniform-circular-motion | <p>A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be :</p> | [{"identifier": "A", "content": "1 : 2"}, {"identifier": "B", "content": "2 : 3"}, {"identifier": "C", "content": "2 : 5"}, {"identifier": "D", "content": "1 : 1"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4a1hbu/cf5132ff-fd3d-4892-a446-59d2b498fbb0/0c58ae90-ac6b-11ed-8b11-cb59760cfd30/file-1le4a1hbv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le4a1hbu/cf5132ff-fd3d-4892-a446-59d2b498fbb0/0c58ae90-ac6b-11ed-8b11-cb59760cfd30... | mcq | jee-main-2023-online-24th-january-evening-shift | 9,553 |
1lgozv6yz | physics | circular-motion | uniform-circular-motion | <p>A vehicle of mass $$200 \mathrm{~kg}$$ is moving along a levelled curved road of radius $$70 \mathrm{~m}$$ with angular velocity of $$0.2 ~\mathrm{rad} / \mathrm{s}$$. The centripetal force acting on the vehicle is:</p> | [{"identifier": "A", "content": "$$560 \\mathrm{~N}$$"}, {"identifier": "B", "content": "$$14 \\mathrm{~N}$$"}, {"identifier": "C", "content": "$$2800 \\mathrm{~N}$$"}, {"identifier": "D", "content": "$$2240 \\mathrm{~N}$$"}] | ["A"] | null | The centripetal force acting on an object moving along a circular path of radius $$r$$ and angular velocity $$\omega$$ is given by:
<br/><br/>
$$F_c=mr\omega^2$$
<br/><br/>
where $$m$$ is the mass of the object.
<br/><br/>
In this problem, the vehicle of mass $$m=200 \mathrm{~kg}$$ is moving along a circular path of ra... | mcq | jee-main-2023-online-13th-april-evening-shift | 9,554 |
1lguxs3kl | physics | circular-motion | uniform-circular-motion | <p>A coin placed on a rotating table just slips when it is placed at a distance of $$1 \mathrm{~cm}$$ from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :</p> | [{"identifier": "A", "content": "1 cm"}, {"identifier": "B", "content": "8 cm"}, {"identifier": "C", "content": "4 cm"}, {"identifier": "D", "content": "2 cm"}] | ["C"] | null | When a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force. Let's denote the mass of the coin as $$m$$, the initial angular velocity as $$\omega_1$$, and the final angular velocity as $$\omega_2$$.
<br/><br/>
Initially, when the... | mcq | jee-main-2023-online-11th-april-morning-shift | 9,555 |
1lh301c4l | physics | circular-motion | uniform-circular-motion | <p>As shown in the figure, a particle is moving with constant speed $$\pi ~\mathrm{m} / \mathrm{s}$$. Considering its motion from $$\mathrm{A}$$ to $$\mathrm{B}$$, the magnitude of the average velocity is :
</p>
<p><img src="data:image/png;base64,UklGRigQAABXRUJQVlA4IBwQAAAQDAGdASoAA0kCP4HA3WQ2Ma6mo9V5AsAwCWlu4XEUhmNwu... | [{"identifier": "A", "content": "$$\\pi ~\\mathrm{m} / \\mathrm{s}$$"}, {"identifier": "B", "content": "$$1.5 \\sqrt{3} \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "C", "content": "$$\\sqrt{3} \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{3} \\mathrm{~m} / \\mathrm{s}$$"}] | ["B"] | null | The given situation is shown below.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh5odxbz/ec1420f6-521b-4334-826d-cab133e5fa6f/950378f0-e894-11ed-8179-292e62481b94/file-1lh5odxc0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh5odxbz/ec1420f6-521b-4334-826d-cab133e5... | mcq | jee-main-2023-online-6th-april-evening-shift | 9,558 |
1lh30s344 | physics | circular-motion | uniform-circular-motion | <p>A child of mass $$5 \mathrm{~kg}$$ is going round a merry-go-round that makes 1 rotation in $$3.14 \mathrm{~s}$$. The radius of the merry-go-round is $$2 \mathrm{~m}$$. The centrifugal force on the child will be</p> | [{"identifier": "A", "content": "50 N"}, {"identifier": "B", "content": "80 N"}, {"identifier": "C", "content": "100 N"}, {"identifier": "D", "content": "40 N"}] | ["D"] | null | <p>To calculate the centrifugal force acting on the child, we need to find the angular velocity of the merry-go-round and then apply the formula for centrifugal force. </p>
<p>The merry-go-round makes 1 rotation in 3.14 seconds, so its angular velocity ($$\omega$$) can be calculated as:</p>
<p>$$\omega = \frac{2\pi}{T}... | mcq | jee-main-2023-online-6th-april-evening-shift | 9,559 |
lsbkstf7 | physics | circular-motion | uniform-circular-motion | A ball of mass $0.5 \mathrm{~kg}$ is attached to a string of length $50 \mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400 \mathrm{~N}$. The maximum possible value of angular velocity of the ball in $\mathrm{rad} / \mathrm{s}$ i... | [{"identifier": "A", "content": "1600"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "40"}, {"identifier": "D", "content": "1000"}] | ["C"] | null | <p>To find the maximum possible angular velocity (ω) of the ball, we need to consider the maximum tension the string can bear without breaking. This tension provides the centripetal force needed to keep the ball in a circular path.</p>
<p>The centripetal force (F<sub>c</sub>) required for circular motion is given by t... | mcq | jee-main-2024-online-1st-february-morning-shift | 9,560 |
lsbla2py | physics | circular-motion | uniform-circular-motion | A particle moving in a circle of radius $\mathrm{R}$ with uniform speed takes time $\mathrm{T}$ to complete one revolution.<br/><br/>
If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4 R$. The angle of projection $\theta$ is then gi... | [{"identifier": "A", "content": "$\\sin ^{-1}\\left[\\frac{2 \\mathrm{gT}^2}{\\pi^2 \\mathrm{R}}\\right]^{\\frac{1}{2}}$"}, {"identifier": "B", "content": "$\\sin ^{-1}\\left[\\frac{\\pi^2 \\mathrm{R}}{2 \\mathrm{gT}^2}\\right]^{\\frac{1}{2}}$"}, {"identifier": "C", "content": "$\\cos ^{-1}\\left[\\frac{\\pi \\mathrm{R... | ["A"] | null | <p>To solve for the angle of projection $\theta$, we will first establish the relationship between the variables given and then derive the formula using kinematics.</p>
<p>The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:</p>
<p>$$ v =... | mcq | jee-main-2024-online-1st-february-morning-shift | 9,561 |
jaoe38c1lsc35sf0 | physics | circular-motion | uniform-circular-motion | <p>A train is moving with a speed of $$12 \mathrm{~m} / \mathrm{s}$$ on rails which are $$1.5 \mathrm{~m}$$ apart. To negotiate a curve radius $$400 \mathrm{~m}$$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $$g=10 \mathrm{~m} / \mathrm{s}^2)$$ :</p> | [{"identifier": "A", "content": "6.0 cm"}, {"identifier": "B", "content": "5.4 cm"}, {"identifier": "C", "content": "4.8 cm"}, {"identifier": "D", "content": "4.2 cm"}] | ["B"] | null | <p>$$\tan \theta=\frac{v^2}{R g}=\frac{12 \times 12}{10 \times 400}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lscudf4i/53002a9a-3b22-4fee-9d31-e8f7de906502/db6ffb20-c64b-11ee-9d8b-f1be86a1b2f3/file-6y3zli1lscudf4j.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/ima... | mcq | jee-main-2024-online-27th-january-morning-shift | 9,562 |
jaoe38c1lse6hrcw | physics | circular-motion | uniform-circular-motion | <p>A coin is placed on a disc. The coefficient of friction between the coin and the disc is $$\mu$$. If the distance of the coin from the center of the disc is $$r$$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\frac{r}{\\mu g}}$$\n"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{\\mu g}{r}}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\mu g}{r}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\mu}{\\sqrt{r g}}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsluc3m3/d6f96f5a-f4c4-4f9b-a47b-e687d115c288/c494a1b0-cb3e-11ee-ad47-a16d1086e690/file-6y3zli1lsluc3m4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsluc3m3/d6f96f5a-f4c4-4f9b-a47b-e687d115c288/c494a1b0-cb3e-11ee... | mcq | jee-main-2024-online-31st-january-morning-shift | 9,563 |
jaoe38c1lsf1k2mp | physics | circular-motion | uniform-circular-motion | <p>If the radius of curvature of the path of two particles of same mass are in the ratio $$3: 4$$, then in order to have constant centripetal force, their velocities will be in the ratio of :</p> | [{"identifier": "A", "content": "$$1: \\sqrt{3}$$\n"}, {"identifier": "B", "content": "$$2: \\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{3}: 2$$\n"}, {"identifier": "D", "content": "$$\\sqrt{3}: 1$$"}] | ["C"] | null | <p>Given $$\mathrm{m}_1=\mathrm{m}_2$$</p>
<p>$$\text { and } \frac{r_1}{r_2}=\frac{3}{4}$$</p>
<p>As centripetal force $$\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}$$</p>
<p>In order to have constant (same in this question) centripetal force</p>
<p>$$\begin{aligned}
& \mathrm{F}_1=\mathrm{F}_2 \\
& \frac{\mathrm{m}_1 ... | mcq | jee-main-2024-online-29th-january-morning-shift | 9,564 |
jaoe38c1lsfln4re | physics | circular-motion | uniform-circular-motion | <p>A stone of mass $$900 \mathrm{~g}$$ is tied to a string and moved in a vertical circle of radius $$1 \mathrm{~m}$$ making $$10 \mathrm{~rpm}$$. The tension in the string, when the stone is at the lowest point is (if $$\pi^2=9.8$$ and $$g=9.8 \mathrm{~m} / \mathrm{s}^2$$) :</p> | [{"identifier": "A", "content": "17.8 N"}, {"identifier": "B", "content": "97 N"}, {"identifier": "C", "content": "9.8 N"}, {"identifier": "D", "content": "8.82 N"}] | ["C"] | null | <p>Given that</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr7p1jk/c47a89d4-e629-47c6-95c2-1f208c4603e0/ee6d4810-ce32-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr7p1jl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr7p1jk/c47a89d4-e629-47c6-95c2-1f208c4603e0/... | mcq | jee-main-2024-online-29th-january-evening-shift | 9,565 |
lv2es3k6 | physics | circular-motion | uniform-circular-motion | <p>A cyclist starts from the point $$P$$ of a circular ground of radius $$2 \mathrm{~km}$$ and travels along its circumference to the point $$\mathrm{S}$$. The displacement of a cyclist is:</p>
<p><img src="data:image/png;base64,UklGRuAPAABXRUJQVlA4INQPAABQFgGdASriAgADP4HA3GW2MS0nIfK4+sAwCWlu/AT4ooOnZ1+ftv6U7uzj/7Zf7dx... | [{"identifier": "A", "content": "$$\\sqrt8$$ km"}, {"identifier": "B", "content": "4 km"}, {"identifier": "C", "content": "6 km "}, {"identifier": "D", "content": "8 km"}] | ["A"] | null | <p>$$\begin{aligned}
\text { Displacement } & =\sqrt{2} R \\
& =\sqrt{8} \mathrm{~km}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift | 9,566 |
lv5gsxza | physics | circular-motion | uniform-circular-motion | <p>A clock has $$75 \mathrm{~cm}, 60 \mathrm{~cm}$$ long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel $$x$$ distance more than the tip of minute hand. The value of $$x$$ in meter is nearly (Take $$\pi=3.14$$) :</p> | [{"identifier": "A", "content": "118.9"}, {"identifier": "B", "content": "140.5"}, {"identifier": "C", "content": "139.4"}, {"identifier": "D", "content": "220.0"}] | ["C"] | null | <p>To determine the distance traveled by the tips of the second hand and the minute hand, we need to calculate the circumference of the circles they make. Let's start by calculating the distances traveled by both hands over a period of 30 minutes.</p>
<p>First, the circumference formula is given by:</p>
<p>$$C = 2\pi... | mcq | jee-main-2024-online-8th-april-morning-shift | 9,567 |
lvb2955q | physics | circular-motion | uniform-circular-motion | <p>A car of $$800 \mathrm{~kg}$$ is taking turn on a banked road of radius $$300 \mathrm{~m}$$ and angle of banking $$30^{\circ}$$. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: $$(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73)$$</p> | [{"identifier": "A", "content": "51.4 m/s"}, {"identifier": "B", "content": "102.8 m/s"}, {"identifier": "C", "content": "70.4 m/s"}, {"identifier": "D", "content": "264 m/s"}] | ["A"] | null | <p>When a car takes a turn on a banked road, the forces involved are the gravitational force, the normal force from the surface, and frictional force (if any). The net force provides the necessary centripetal force for the circular motion. The angle of banking and static friction contribute to the maximum speed the car... | mcq | jee-main-2024-online-6th-april-evening-shift | 9,569 |
sfLB2JRpKsBNbp75 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | Consider telecommunication through optical fibres. Which of the following statements is <b>not</b> true? | [{"identifier": "A", "content": "Optical fibres can be of graded refractive index"}, {"identifier": "B", "content": "Optical fibres are subject to electromagnetic interference from outside "}, {"identifier": "C", "content": "Optical fibres have extremely low transmission loss "}, {"identifier": "D", "content": "Optical... | ["B"] | null | Optical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference. | mcq | aieee-2003 | 9,570 |
wSTcKrPWSaJwXk7o | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | This question has Statement - $$1$$ and Statement - $$2$$. Of the four choices given after the statements, choose the one that best describes the two statements.
<br/><b>Statement -$$1$$ :</b> Sky wave signals are used for long distance radio communication. These signals are in general, less stable then ground wave sig... | [{"identifier": "A", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is the correct explanation of Statement - $$1$$."}, {"identifier": "B", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is not the correct explanation of Statement - $$1$$."}, {"i... | ["B"] | null | For long distance communication, sky wave signals are used.
<br><br>Also, the state of ionosphere varies every time. So, both statements are correct. | mcq | aieee-2011 | 9,571 |
YJfS2N6BUi79xvE7 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A radar has a power of $$1kW$$ and is operating at a frequency of $$10$$ $$GHz.$$ It is located on a mountain top of height $$500$$ $$m.$$ The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth $$ = 6.4 \times {10^6}m$$) is : | [{"identifier": "A", "content": "$$80$$ $$km$$ "}, {"identifier": "B", "content": "$$16$$ $$km$$ "}, {"identifier": "C", "content": "$$40$$ $$km$$ "}, {"identifier": "D", "content": "$$64$$ $$km$$ "}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263539/exam_images/txcxddg3y9nz52bi3dio.webp" loading="lazy" alt="AIEEE 2012 Physics - Communication Systems Question 99 English Explanation">
<br><br>Let $$d$$ is the maximum distance, upto it the objects
<br><br>From $$\Delta AOC... | mcq | aieee-2012 | 9,572 |
UhPPnhFnvgkUJB08YfxUy | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A signal is to be transmitted through a wave of wavelength $$\lambda $$, using a linear antenna. The length l of the antenna and effective power radiated P<sub>eff</sub> will be given respectively as :
<br/><br/>(K is a constant of proportionality)
| [{"identifier": "A", "content": "$$\\lambda $$, P<sub>eff</sub> = K $$\\left( {{1 \\over \\lambda }} \\right)$$<sup>2</sup>"}, {"identifier": "B", "content": "$${{\\lambda \\over 8}}$$, P<sub>eff</sub> = K $$\\left( {{1 \\over \\lambda }} \\right)$$"}, {"identifier": "C", "content": "$${{\\lambda \\over 16}}$$, P<sub... | ["A"] | null | <p>We know that for a linear antenna of length l, (i) it's length of antenna is comparable to the wavelength of the signal and (ii) the power radiated is proportional to 1/$$\lambda$$<sup>2</sup>. That is, the power radiated for the antenna increases with decreasing $$\lambda$$ thereby increasing the frequency.</p>
<p>... | mcq | jee-main-2017-online-9th-april-morning-slot | 9,573 |
lbc1eczRhGSFNJG8 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized
for transmission. How many telephonic channels can be transmitted simultaneously if each channel
requires a bandwidth of 5 kHz ? | [{"identifier": "A", "content": "2 $$\\times$$ 10<sup>6</sup>"}, {"identifier": "B", "content": "2 $$\\times$$ 10<sup>3</sup>"}, {"identifier": "C", "content": "2 $$\\times$$ 10<sup>5</sup>"}, {"identifier": "D", "content": "2 $$\\times$$ 10<sup>4</sup>"}] | ["C"] | null | <p>The carrier frequency is 10 GHz, and only 10% of it is utilized for transmission. Therefore, the bandwidth used for transmission is:</p>
<p>$ \text{Bandwidth} = 10\% \times 10 \text{ GHz} = 0.1 \times 10 \text{ GHz} = 1 \text{ GHz}$</p>
<p>We should convert this to kHz because the channel bandwidth is given in kHz:<... | mcq | jee-main-2018-offline | 9,574 |
xWfKfyF8DZZhtEBdzWB9U | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower is LOS (Line of Sight) mode ? (Given : radius of earth = 6.4 × 10<sup>6</sup> m). | [{"identifier": "A", "content": "40 km"}, {"identifier": "B", "content": "65 km"}, {"identifier": "C", "content": "48 km"}, {"identifier": "D", "content": "80 km"}] | ["B"] | null | Maximum distance upto which signal can be broadcasted is
<br><br>d<sub>max</sub> = $$\sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$
<br><br>where h<sub>T</sub> and h<sub>R</sub> are heights of transmitter tower and height of reserver respectively. Putting all values -
<br><br>d<sub>max</sub> = $$\sqrt {2 \times 6.4 \times 106}... | mcq | jee-main-2019-online-10th-january-morning-slot | 9,576 |
dLxH6Mvqz41KHVUUnF6VE | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | To double the covering range of a TV transmittion tower, its height should be multiplied by : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 2 }}$$"}] | ["A"] | null | <p>Range of TV transmitting tower $$d = \sqrt {2hR} $$ where, h is the height of the transmission tower. When range is doubled,</p>
<p>$$\therefore$$ $$2d = 2\sqrt {2hR} = \sqrt {2(4h)R} $$</p>
<p>So, height must be multiplied with 4.</p> | mcq | jee-main-2019-online-12th-january-evening-slot | 9,577 |
o9kDJjw1iiuS0Lc5957MG | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | The wavelength of the carrier waves in a
modern optical fiber communication network
is close to : | [{"identifier": "A", "content": "1500 nm"}, {"identifier": "B", "content": "2400 nm"}, {"identifier": "C", "content": "600 nm"}, {"identifier": "D", "content": "900 nm"}] | ["A"] | null | Wavelength of carrier waves in modern optical
fiber communication is most widely used near
about 1500 nm. | mcq | jee-main-2019-online-8th-april-morning-slot | 9,578 |
AbyExxF0GdOnsWyadlFpp | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | In a line of sight radio communication, a
distance of about 50 km is kept between the
transmitting and receiving antennas. If the
height of the receiving antenna is 70m, then the
minimum height of the transmitting antenna
should be :<br/>
(Radius of the Earth = 6.4 × 10<sup>6</sup> m). | [{"identifier": "A", "content": "40 m"}, {"identifier": "B", "content": "20 m"}, {"identifier": "C", "content": "51 m"}, {"identifier": "D", "content": "32 m"}] | ["D"] | null | Range = $$\sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$<br><br>
$$50 \times {10^3} = \sqrt {2 \times 6400 \times {{10}^3} \times {h_T}} + \sqrt {2 \times 6400 \times {{10}^3} \times 70} $$<br><br>
By solving h<sub>T</sub> = 32 m. | mcq | jee-main-2019-online-8th-april-evening-slot | 9,579 |
5xydr3OdDN6Ihpn7iIZCR | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | The physical sizes of the transmitter and
receiver antenna in a communication system
are :- | [{"identifier": "A", "content": "proportional to carrier frequency"}, {"identifier": "B", "content": "inversely proportional to carrier frequency"}, {"identifier": "C", "content": "inversely proportional to modulation\nfrequency"}, {"identifier": "D", "content": "independent of both carrier and modulation\nfrequency"}] | ["B"] | null | The physical size of antenna of receiver and transmitter both are inversely proportional
to carrier frequency. | mcq | jee-main-2019-online-9th-april-evening-slot | 9,580 |
JV5dotDpVil9I3VlYK18hoxe66ijvzttrrz | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | Given below in the the left column are different
modes of communication using the kinds of
waves given the right column.
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width: 100%}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:... | [{"identifier": "A", "content": "A-Q, B-S, C-R, D-P"}, {"identifier": "B", "content": "A-Q, B-S, C-P, D-R"}, {"identifier": "C", "content": "A-S, B-Q, C-R, D-P"}, {"identifier": "D", "content": "A-R, B-P, C-S, D-Q"}] | ["B"] | null | Optical Fibre Communication – Infrared Light<br><br/>
Radar – Radio Waves<br><br/>
Sonar – Ultrasound<br><br/>
Mobile Phones – Microwaves | mcq | jee-main-2019-online-10th-april-morning-slot | 9,581 |
9kZKKAhwaqrr7MEKRL1klrp0euf | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is $$-$$5 dB per km and cable length is 20 km. The power received at receiver is 10<sup>$$-$$x</sup> W. The value of x is ________.<br/><br/>[Gain in $$dB = 10{\log _{10}}\left( {{{{P_o}} \over {{P_i}}}} \right)$$] | [] | null | 8 | Given, power of transmitted signal, P<sub>i</sub> = 0.1 kW = 0.1 $$\times$$ 10<sup>3</sup>W = 10<sup>2</sup> W<br/><br/>Rate of attenuation, R = $$-$$5 dB/km<br/><br/>Length of cable, l = 20 km<br/><br/>Power received at receiver, P<sub>x</sub> = 10<sup>$$-$$x</sup> W<br/><br/>Total loss, $$\beta$$ = R $$\times$$ l = $... | integer | jee-main-2021-online-24th-february-evening-slot | 9,582 |
UvVPwu9BXBwe5poWCQ1kmhnbkz9 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 m. The wavelength (in meter) of the signal transmitted by this antenna would be : | [{"identifier": "A", "content": "200"}, {"identifier": "B", "content": "400"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "300"}] | ["C"] | null | Given that, height of peak of antenna : H = 25 m.<br><br>As, we know that<br><br>$$\lambda$$ = 4H<br><br>$$ \therefore $$ $$\lambda$$ = 4 $$\times$$ 25<br><br>$$ \Rightarrow $$ $$\lambda$$ = 100 m | mcq | jee-main-2021-online-16th-march-morning-shift | 9,583 |
cbu15v1cs1dnU7avgZ1kmior527 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | Two identical antennas mounted on identical towers are separated from each other by a distance of 45 km. What should nearly be the minimum height of receiving antenna to receive the signals in line of sight? (Assume radius of earth is 6400 km) | [{"identifier": "A", "content": "158.2 m"}, {"identifier": "B", "content": "79.1 m"}, {"identifier": "C", "content": "19.77 m"}, {"identifier": "D", "content": "39.55 m"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264554/exam_images/me7nyal89af3qde93tyd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Physics - Communication Systems Question 65 English Explanation"><br><br>... | mcq | jee-main-2021-online-16th-march-evening-shift | 9,584 |
rKD00vaanRTia79fZr1kmj3wxf0 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | For VHF signal broadcasting, ___________ km<sup>2</sup> of maximum service area will be covered by an antenna tower of height 30 m, if the receiving antenna is placed at ground. Let radius of the earth be 6400 km. (Round off to the Nearest Integer) (Take $$\pi$$ as 3.14) | [] | null | 1206 | $$d = \sqrt {2hR} $$<br><br>area = $$\pi$$d<sup>2</sup><br><br>Area = $$\pi$$(2hR) = 3.14 $$\times$$ 2 $$\times$$ 30 $$\times$$ 6400 $$\times$$ 10<sup>3</sup> . m<sup>2</sup><br><br>= 1205.76 km<sup>2</sup><br><br>$$ \approx $$ 1206 km<sup>2</sup> | integer | jee-main-2021-online-17th-march-morning-shift | 9,585 |
EsDEDVxBXUUytVfVJg1kmkqve23 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | Match List - I with List - II.<br/><br/>List - I<br/><br/>(a) 10 km height over earth's surface<br/><br/>(b) 70 km height over earth's surface<br/><br/>(c) 180 km height over earth's surface<br/><br/>(d) 270 km height over earth's surface<br/><br/>List - II<br/><br/>(i) Thermosphere<br/><br/>(ii) Mesosphere<br/><br/>(i... | [{"identifier": "A", "content": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)"}, {"identifier": "B", "content": "(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)"}, {"identifier": "C", "content": "(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)"}, {"identifier": "D", "content": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)"}] | ["A"] | null | <b>Troposphere :</b>
<br>The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high
(6 to 9 miles).
<br><br><b>Stratosphere :</b>
<br>The stratosphere starts just above the troposphere and extends to 50 kilometers (31 miles)
high.
<br><br><b>Mesosphere :</b>
<br>The mesosphere starts just above... | mcq | jee-main-2021-online-18th-march-morning-shift | 9,586 |
1krsvs6g7 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | What should be the height of transmitting antenna and the population covered if the television telecast is to cover a radius of 150 km? The average population density around the tower is 2000/km<sup>2</sup> and the value of R<sub>e</sub> = 6.5 $$\times$$ 10<sup>6</sup> m. | [{"identifier": "A", "content": "Height = 1241 m<br><br>Population covered = 7 $$\\times$$ 10<sup>5</sup>"}, {"identifier": "B", "content": "Height = 1731 m<br><br>Population covered = 1413 $$\\times$$ 10<sup>5</sup>"}, {"identifier": "C", "content": "Height = 1800 m<br><br>Population covered = 1413 $$\\times$$ 10<sup>... | ["B"] | null | $$d = \sqrt {2{R_e}h} $$<br><br>Area covered = $$\pi$$d<sup>2</sup><br><br>$$h = {{{d^2}} \over {2{R_e}}}$$ = 1731 m<br><br>Population covered = 2000 $$\times$$ $$\pi$$(d<sup>2</sup>) | mcq | jee-main-2021-online-22th-july-evening-shift | 9,588 |
1ktbqwwqo | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A transmitting antenna at top of a tower has a height of 50 m and the height of receiving antenna is 80 m. What is range of communication for Line of Sight (LoS) mode?<br/><br/>[use radius of earth = 6400 km] | [{"identifier": "A", "content": "45.5 km"}, {"identifier": "B", "content": "80.2 km"}, {"identifier": "C", "content": "144.1 km"}, {"identifier": "D", "content": "57.28 km"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266908/exam_images/r83onjxtatbp4zew98ss.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Evening Shift Physics - Communication Systems Question 53 English Explanation"> <br><b... | mcq | jee-main-2021-online-26th-august-evening-shift | 9,589 |
1kte7oij3 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is ................. km. | [] | null | 224 | $${d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$<br><br>$${d_m} = \left( {\sqrt {2 \times 6400 \times {{10}^3} \times 320} + \sqrt {2 \times 6400 \times {{10}^3} \times 2000} } \right)$$m<br><br>d<sub>m</sub> = 224 km | integer | jee-main-2021-online-27th-august-morning-shift | 9,590 |
1ktfluwou | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km? | [{"identifier": "A", "content": "37.8 m"}, {"identifier": "B", "content": "605 m"}, {"identifier": "C", "content": "75.6 m"}, {"identifier": "D", "content": "302 m"}] | ["B"] | null | h : height of antenna<br><br>$$\lambda$$ : wavelength of signal<br><br>h < $$\lambda$$<br><br>$$\lambda$$ > h<br><br>$$\lambda$$ > 400 m | mcq | jee-main-2021-online-27th-august-evening-shift | 9,591 |
1kth5ukwj | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is ___________ km. (Take radius of Earth = 6400 km) | [] | null | 64 | h<sub>T</sub> = h<sub>R</sub> = 160 ........ (i)<br><br>$$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$<br><br>$$d = \sqrt {2R} \left[ {\sqrt {{h_T}} + \sqrt {{h_R}} } \right]$$<br><br>$$d = \sqrt {2R} \left[ {\sqrt x + \sqrt {160 - x} } \right]$$<br><br>$${{d(d)} \over {dx}} = 0$$<br><br>$${1 \over {2\sqrt x }} + {{1( -... | integer | jee-main-2021-online-31st-august-morning-shift | 9,592 |
1l569wd1g | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>Match List-I with List-II</p>
<p><style type="text/css">
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.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-I, B-II, C-III, D-IV"}, {"identifier": "B", "content": "A-IV, B-III, C-I, D-II"}, {"identifier": "C", "content": "A-IV, B-III, C-II, D-I"}, {"identifier": "D", "content": "A-I, B-II, C-IV, D-III"}] | ["C"] | null | <p>Television signal $$\Rightarrow$$ 6 MHz</p>
<p>Radio signal $$\Rightarrow$$ 2 MHz</p>
<p>High Quality music $$\Rightarrow$$ 20 kHz</p>
<p>Human speech $$\Rightarrow$$ 3 kHz</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 9,594 |
1l56vkbk3 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>We do not transmit low frequency signal to long distances because-</p>
<p>(a) The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength.</p>
<p>(b) Effective power radiated by a long wavelength baseband signal would be high.</p>
<p>(c) We want to avo... | [{"identifier": "A", "content": "All statements are true"}, {"identifier": "B", "content": "(a), (b) and (c) are true only"}, {"identifier": "C", "content": "(a), (c) and (d) are true only"}, {"identifier": "D", "content": "(b), (c) and (d) are true only"}] | ["C"] | null | <p>For longer wavelength, size of antenna would increase. Also, mixing of signals needs to be avoided.</p>
<p>Also, we can use modulation to send low frequency signal by superimposing them with high frequency signals.</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 9,595 |
1l57qw3sm | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is, 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is K$$\sqrt5$$ $$\times$$ 10<sup>2</sup> m. The value of K is ___________.</p>
<p>(Assume radius of Earth is 64 $$\times$... | [] | null | 192 | <p>$$d = \sqrt {2{h_t}{R_e}} + \sqrt {2 \times {h_R}{R_e}} $$</p>
<p>$$ = \sqrt {2 \times 25 \times 64 \times {{10}^5}} + \sqrt {2 \times 49 \times 64 \times {{10}^5}} $$</p>
<p>$$ = 8000\sqrt 5 + 11200\sqrt 5 $$ m</p>
<p>$$ = 19200\sqrt 5 $$ m</p>
<p>$$ = 192\sqrt 5 + {10^2}$$ m</p> | integer | jee-main-2022-online-27th-june-morning-shift | 9,596 |
1l5al1g80 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>A signal of 100 THz frequency can be transmitted with maximum efficiency by :</p> | [{"identifier": "A", "content": "Coaxial cable"}, {"identifier": "B", "content": "Optical fibre"}, {"identifier": "C", "content": "Twisted pair of copper wires"}, {"identifier": "D", "content": "Water"}] | ["B"] | null | <p>Optical fibres supports frequency of electromagnetic waves in the range 10<sup>14</sup> Hz to 10<sup>15</sup> Hz.</p> | mcq | jee-main-2022-online-25th-june-morning-shift | 9,597 |
1l5bcdl1x | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of __________ GHz.</p>
<p>(Given $$\mu$$<sub>r</sub> = 1 for dielectric medium)</p> | [] | null | 6 | <p>We know that v = f$$\lambda$$</p>
<p>Putting the values,</p>
<p>$${{3 \times {{10}^8}} \over {\sqrt {6.25} }} = f \times 20 \times {10^{ - 3}}$$</p>
<p>$$ \Rightarrow f = 6 \times {10^9}$$ Hz</p> | integer | jee-main-2022-online-24th-june-evening-shift | 9,598 |
1l6e0fbcd | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>The required height of a TV tower which can cover the population of $$6.03$$ lakh is $$h$$. If the average population density is 100 per square $$\mathrm{km}$$ and the radius of earth is $$6400 \mathrm{~km}$$, then the value of $$h$$ will be _____________ $$m$$.</p> | [] | null | 150 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6tc4h2y/fe8dff0e-e551-4dae-8c49-554ce7caec00/0f0bcda0-1bd1-11ed-bd9f-71db3a4811b2/file-1l6tc4h2z.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6tc4h2y/fe8dff0e-e551-4dae-8c49-554ce7caec00/0f0bcda0-1bd1-11ed-bd9f-71db3a4811b2... | integer | jee-main-2022-online-25th-july-morning-shift | 9,599 |
1ldofb88n | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>Which of the following frequencies does not belong to FM broadcast ?</p> | [{"identifier": "A", "content": "$$106 ~\\mathrm{MHz}$$"}, {"identifier": "B", "content": "$$89 ~\\mathrm{MHz}$$"}, {"identifier": "C", "content": "$$64 ~\\mathrm{MHz}$$"}, {"identifier": "D", "content": "$$99 ~\\mathrm{MHz}$$"}] | ["C"] | null | FM broadcast range is 88MHz to 108MHz. | mcq | jee-main-2023-online-1st-february-morning-shift | 9,601 |
1ldspsli2 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>If the height of transmitting and receiving antennas are 80 m each, the maximum line of sight distance will be:</p>
<p>Given : Earth's radius $$=6.4\times10^6$$ m</p> | [{"identifier": "A", "content": "28 km"}, {"identifier": "B", "content": "64 km"}, {"identifier": "C", "content": "32 km"}, {"identifier": "D", "content": "36 km"}] | ["B"] | null | Maximum line of sight distance between two antennas, $\mathrm{d}_{\mathrm{M}}=\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{R} \cdot \mathrm{h}_{\mathrm{R}}}$
<br/><br/>
$\mathrm{d}_{\mathrm{M}}=2 \times \sqrt{2 \times 6.4 \times 10^{6} \times 80}=64 \mathrm{~km}$ | mcq | jee-main-2023-online-29th-january-morning-shift | 9,602 |
1ldtxrfri | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>Match List I with List II</p>
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.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-III, B-IV, C-II, D-I"}, {"identifier": "B", "content": "A-III, B-II, C-I, D-IV"}, {"identifier": "C", "content": "A-I, B-IV, C-III, D-II"}, {"identifier": "D", "content": "A-I, B-II, C-IV, D-III"}] | ["A"] | null | $\rightarrow 10 \mathrm{~km}$ over Earth's surface - Troposphere
<br/><br/>
$\rightarrow 100 \mathrm{~km}$ over Earth's surface - E-part of stratosphere
<br/><br/>
$\rightarrow 300 \mathrm{~km}$ over Earth's surface $-\mathrm{F}_{2}$-part of thermosphere
<br/><br/>
$\rightarrow$ 65-75 km over Earth's surface - D-part o... | mcq | jee-main-2023-online-25th-january-evening-shift | 9,603 |
lgny32wf | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | The height of transmitting antenna is $180 \mathrm{~m}$ and the height of the receiving antenna is $245 \mathrm{~m}$. The maximum distance between them for satisfactory communication in line of sight will be :<br/><br/>(given $\mathrm{R}=6400 \mathrm{~km})$ | [{"identifier": "A", "content": "$104 \\mathrm{~km}$"}, {"identifier": "B", "content": "$56 \\mathrm{~km}$"}, {"identifier": "C", "content": "$48 \\mathrm{~km}$"}, {"identifier": "D", "content": "$96 \\mathrm{~km}$"}] | ["A"] | null |
$$
d_{\max} = \sqrt{2Rh_t} + \sqrt{2Rh_r}
$$
<br/><br/>
where $d_{\max}$ is the maximum distance between the antennas, $R$ is the radius of the earth, $h_t$ is the height of the transmitting antenna above the earth's surface, and $h_r$ is the height of the receiving antenna above the earth's surface.
<br/><br/>
Substi... | mcq | jee-main-2023-online-15th-april-morning-shift | 9,605 |
1lgoztiqi | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>To radiate EM signal of wavelength $$\lambda$$ with high efficiency, the antennas should have a minimum size equal to:</p> | [{"identifier": "A", "content": "$$\\frac{\\lambda}{2}$$"}, {"identifier": "B", "content": "$$\\lambda$$"}, {"identifier": "C", "content": "$$\\frac{\\lambda}{4}$$"}, {"identifier": "D", "content": "$$2 \\lambda$$"}] | ["C"] | null | The minimum length of an antenna to efficiently radiate an electromagnetic wave is one quarter of the wavelength of the signal, which is commonly known as a quarter-wave antenna.
<br/><br/>
This is because when the length of the antenna is equal to a quarter of the wavelength of the signal, the voltage and current at ... | mcq | jee-main-2023-online-13th-april-evening-shift | 9,606 |
1lgq18bgf | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>Match List - I with List - II</p>
<p><style type="text/css">
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.tg th{border-color:black;border-style:soli... | [{"identifier": "A", "content": "A - II, B - IV, C - III, D - I"}, {"identifier": "B", "content": "A - II, B-I, C - IV, D - III"}, {"identifier": "C", "content": "A - II, B - IV, C - I, D - III"}, {"identifier": "D", "content": "A - III, B - IV, C - I, D - II"}] | ["C"] | null | Let's match the layers of the atmosphere (List - I) with their approximate heights over the Earth's surface (List - II):
<br/><br/>
(A) $$\mathrm{F_1}$$ - Layer: This layer is part of the ionosphere, which is located within the thermosphere. The $$\mathrm{F_1}$$ layer is at an approximate height of 170 - 190 km. So, A ... | mcq | jee-main-2023-online-13th-april-morning-shift | 9,607 |
1lgsx1izi | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>In satellite communication, the uplink frequency band used is :</p> | [{"identifier": "A", "content": "$$3.7-4.2 \\mathrm{~GHz}$$"}, {"identifier": "B", "content": "$$5.925-6.425 \\mathrm{~GHz}$$"}, {"identifier": "C", "content": "$$76-88 \\mathrm{~MHz}$$"}, {"identifier": "D", "content": "$$420-890 \\mathrm{~MHz}$$"}] | ["B"] | null | In satellite communication, the uplink frequency band refers to the range of frequencies used to transmit signals from the Earth to the satellite. The specific frequency bands used for uplink communication can vary depending on the type of satellite and the service it provides. However, one of the most common frequency... | mcq | jee-main-2023-online-11th-april-evening-shift | 9,608 |
1lguy5qu6 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>A transmitting antenna is kept on the surface of the earth. The minimum height of receiving antenna required to receive the signal in line of sight at $$4 \mathrm{~km}$$ distance from it is $$x \times 10^{-2} \mathrm{~m}$$. The value of $$x$$ is __________.</p>
<p>(Let, radius of earth $$\mathrm{R}=6400 \mathrm{~km}... | [{"identifier": "A", "content": "1.25"}, {"identifier": "B", "content": "125"}, {"identifier": "C", "content": "1250"}, {"identifier": "D", "content": "12.5"}] | ["B"] | null | <p>To determine the minimum height of the receiving antenna required to receive the signal in line of sight at a distance of 4 km from the transmitting antenna, we need to consider the curvature of the Earth. Let's denote the height of the receiving antenna as $$h$$ and the distance between the antennas as $$d = 4 ... | mcq | jee-main-2023-online-11th-april-morning-shift | 9,609 |
1lgyqlpcj | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>The power radiated from a linear antenna of length $$l$$ is proportional to</p>
<p>(Given, $$\lambda=$$ Wavelength of wave):</p> | [{"identifier": "A", "content": "$$\\left(\\frac{l}{\\lambda}\\right)^{2}$$"}, {"identifier": "B", "content": "$$\\frac{l}{\\lambda^{2}}$$"}, {"identifier": "C", "content": "$$\\frac{l^{2}}{\\lambda}$$"}, {"identifier": "D", "content": "$$\\frac{l}{\\lambda}$$"}] | ["A"] | null | The power radiated by a linear antenna is typically determined by its physical length relative to the wavelength of the signal it is transmitting or receiving. The power radiated from a linear antenna of length 'l' is proportional to the square of the ratio of its length to the wavelength, represented as $(\frac{l}{\la... | mcq | jee-main-2023-online-8th-april-evening-shift | 9,610 |
1lh023yn3 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>A TV transmitting antenna is $$98 \mathrm{~m}$$ high and the receiving antenna is at the ground level. If the radius of the earth is $$6400 \mathrm{~km}$$, the surface area covered by the transmitting antenna is approximately:</p> | [{"identifier": "A", "content": "$$4868 \\mathrm{~km}^{2}$$"}, {"identifier": "B", "content": "$$1240 \\mathrm{~km}^{2}$$"}, {"identifier": "C", "content": "$$3942 \\mathrm{~km}^{2}$$"}, {"identifier": "D", "content": "$$1549 \\mathrm{~km}^{2}$$"}] | ["C"] | null | <p>The range (or distance to the horizon) for line-of-sight communication is given by the formula:</p>
<p>$d = \sqrt{2hR}$,</p>
<p>where:</p>
<ul>
<li>$d$ is the range,</li>
<li>$h$ is the height of the antenna, and</li>
<li>$R$ is the radius of the Earth.</li>
</ul>
<p>Given that $h = 98 \, \text{m} = 98 \times 10^{-3... | mcq | jee-main-2023-online-8th-april-morning-shift | 9,611 |
1lh24v2j5 | physics | communication-systems | elements-of-communication-system-&-propagation-of-em-wave | <p>By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by $$21 \%$$ ?</p> | [{"identifier": "A", "content": "12%"}, {"identifier": "B", "content": "15%"}, {"identifier": "C", "content": "10%"}, {"identifier": "D", "content": "14%"}] | ["C"] | null | <p>The range of transmission (R) of a TV tower (or any radio tower) depends on the height of the tower (h). This relationship is given by the formula:</p>
<p>$$
R = \sqrt{2hR_E}
$$</p>
<p>where $R_E$ is the radius of the Earth. This formula comes from the geometry of the situation, considering the curvature of the Eart... | mcq | jee-main-2023-online-6th-april-morning-shift | 9,612 |
xG2RNOrnUQUCyg2v | physics | communication-systems | modulation-&-demodulation | A signal of $$5$$ $$kHz$$ frequency is amplitude modulated on a carrier wave of frequency $$2$$ $$MHz.$$ The frequencies of the resultant signal is/are : | [{"identifier": "A", "content": "$$2005$$ $$kHz, 2000$$ $$kHz$$ and $$1995$$ $$kHz$$"}, {"identifier": "B", "content": "$$2000$$ $$kHz$$ and $$1995$$ $$kHz$$ "}, {"identifier": "C", "content": "$$2$$ $$MHz$$ only "}, {"identifier": "D", "content": "$$2005$$ $$kHz$$ and $$1995$$ $$kHz$$"}] | ["A"] | null | Amplitude modulated wave consists of three frequencies are
<br><br>$${\omega _c} + {\omega _m},\,\omega ,{\omega _c} - {\omega _m}$$
<br><br>i.e. $$2005$$ $$KHz,$$ $$2000KHz,$$ $$1995$$ $$KHz$$ | mcq | jee-main-2015-offline | 9,613 |
CO61bwbkzmBMahH6 | physics | communication-systems | modulation-&-demodulation | Choose the correct statement : | [{"identifier": "A", "content": "In amplitude modulation the amplitude of the high frequency carrier wave is make to vary in proportion to the amplitude of the audio signal. "}, {"identifier": "B", "content": "In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the ... | ["A"] | null | In amplitude modulation, the amplitude of the high frequency carrier wave made to vary in proportional to the amplitude of audio signal.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265501/exam_images/xkrsjni0xrpnrah1p5cq.webp" loading="lazy" alt="JEE Main 2016 (Offlin... | mcq | jee-main-2016-offline | 9,614 |
11D8X5Nnuug2dQM4CjUKT | physics | communication-systems | modulation-&-demodulation | An audio signal consists of two distinct sounds : one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwid... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>To calculate the ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech, we need to consider the frequency range covered by both signals.</p>
<p>For the human speech signal, the frequency band is from 200 Hz to 2700 Hz. Therefore,... | mcq | jee-main-2016-online-9th-april-morning-slot | 9,616 |
Z1o1XZ0inG8g4ug4ExWzq | physics | communication-systems | modulation-&-demodulation | A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement. | [{"identifier": "A", "content": "Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz"}, {"identifier": "B", "content": "Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz\n"}, {"identifier": "C", "content": "Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 k... | ["D"] | null | Modulation index (m) = $${{{V_m}} \over {{V_0}}} = {5 \over {25}} = 0.2$$
<br><br>Given, carrier wave,
<br><br>F<sub>c</sub> = 1.2 $$ \times $$ 10<sup>6</sup> Hz = 1200 kHz
<br><br>and frequency of modulate wave,
<br><br>F<sub>m</sub> = 20 kHz
<br><br>$$\therefore\,\,\,$$ Side frequency bands are
<br><br>F<sub>1</sub... | mcq | jee-main-2017-online-8th-april-morning-slot | 9,617 |
qSPuzFWLll3PAk0IKi4ns | physics | communication-systems | modulation-&-demodulation | A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of 80% will be : | [{"identifier": "A", "content": "7 V"}, {"identifier": "B", "content": "28 V"}, {"identifier": "C", "content": "11.2 V"}, {"identifier": "D", "content": "22.4 V"}] | ["C"] | null | <p>Amplitude of carrier wave is A<sub>c</sub> = 14 V</p>
<p>Modulation Index is $$\mu$$ = 80% = 0.80</p>
<p>Amplitude of modulating signal is given by</p>
<p>$$\mu = {{{A_m}} \over {{A_c}}} \Rightarrow {A_m} = \mu {A_c} \Rightarrow {A_m} = 0.8 \times 14 = 11.2\,V$$</p> | mcq | jee-main-2018-online-16th-april-morning-slot | 9,619 |
Si5UroIW6vHo2aUr5JFxv | physics | communication-systems | modulation-&-demodulation | The number of amplitude modulted broadcast stations that can be accomodated in a $$300$$ $$kHz$$ band width for the highest modulating frequency $$15$$ $$kHz$$ will be : | [{"identifier": "A", "content": "$$20$$"}, {"identifier": "B", "content": "$$15$$ "}, {"identifier": "C", "content": "$$10$$ "}, {"identifier": "D", "content": "$$8$$ "}] | ["C"] | null | Given, modulating frequency = 15 kHz
<br><br>$$\therefore\,\,\,\,$$ Bandwidth of one channel = 2 $$ \times $$ 15 = 30 kHz
<br><br>$$\therefore\,\,\,\,$$ No of stations = $${{300} \over {30}}$$ = 10 | mcq | jee-main-2018-online-15th-april-morning-slot | 9,620 |
tZkq2UIc2P8evijYkrIEo | physics | communication-systems | modulation-&-demodulation | The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot ? | [{"identifier": "A", "content": "2900 kHz"}, {"identifier": "B", "content": "2750 kHz"}, {"identifier": "C", "content": "2250 kHz"}, {"identifier": "D", "content": "2000 kz"}] | ["D"] | null | f<sub>carrier</sub> = $${{250} \over {0.1}}$$ = 2500 KHZ
<br><br>$$ \therefore $$ Range of signal = 2250 Hz to 2750 Hz
<br><br>Now check all options : for 2000 KHZ
<br><br>f<sub>mod</sub> = 200 Hz
<br><br>$$ \therefore $$ Range = 1800 KHZ to 2200 KHZ | mcq | jee-main-2019-online-10th-january-evening-slot | 9,622 |
SzptE5xITq9Xr6QNHsSrd | physics | communication-systems | modulation-&-demodulation | An amplitude modulated signal is plotted below :
<br/><br/><img src="data:image/png;base64,UklGRqQZAABXRUJQVlA4IJgZAACQfQCdASrUAeUAPm00l0ekIyIhJFHrIIANiWlu4XNBG/Mz8J/2v1ZeDP4n7duzb8je4n9t9uvKv19anHx/69flf8L+NPz3/jf9t4a++bUC/MP5t/uN6B2XzBfZ77b/wPub9GT9+/vPqJ+Yf1z/W/bh9gH8Z/mH+y9Pf+D4bHy//XewH/Gf6L/xv7R/hv11+l3+i/9H+R/x... | [{"identifier": "A", "content": "(1 + 9sin (2$$\\pi $$ $$ \\times $$ 10<sup>4</sup> t)) sin(2.5$$\\pi $$ $$ \\times $$ 10<sup>5</sup>t) V"}, {"identifier": "B", "content": "(9 + sin (4$$\\pi $$ $$ \\times $$ 10<sup>4</sup> t)) sin(5$$\\pi $$ $$ \\times $$ 10<sup>5</sup>t) V"}, {"identifier": "C", "content": "(9 + sin (... | ["C"] | null | Analysis of graph says
<br><br>(1) Amplitude varies as 8 $$-$$ 10 V or 9 $$ \pm $$ 1
<br><br>(2) Two time period as
<br><br>100 $$\mu $$s (signal wave) & 8 $$\mu $$s (carrier wave)
<br><br>Hence signal is $$\left[ {9 \pm 1\sin \left( {{{2\pi t} \over {{T_1}}}} \right)} \right]\sin \left( {{{2... | mcq | jee-main-2019-online-11th-january-evening-slot | 9,624 |
EcjLKq094KTSGZyr2M7rP | physics | communication-systems | modulation-&-demodulation | A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index ?
| [{"identifier": "A", "content": "0.5 "}, {"identifier": "B", "content": "0.6"}, {"identifier": "C", "content": "0.4"}, {"identifier": "D", "content": "0.3"}] | ["B"] | null | E<sub>m</sub> + E<sub>c</sub> = 160
<br><br>E<sub>m</sub> + 100 = 160
<br><br>E<sub>m</sub> = 60
<br><br>$$\mu = {{{E_m}} \over {{E_C}}} = {{60} \over {100}}$$
<br><br>$$\mu $$ = 0.6 | mcq | jee-main-2019-online-12th-january-morning-slot | 9,625 |
nmzSejiMMpSYBUCfgarGT | physics | communication-systems | modulation-&-demodulation | A signal Acoswt is transmitted using v<sub>0</sub> sin $$\omega $$<sub>0</sub>t
as carrier wave. The correct amplitude
modulated (AM) signal is | [{"identifier": "A", "content": "v<sub>0</sub> sin $$\\omega $$<sub>0</sub>t + Acos$$\\omega $$t"}, {"identifier": "B", "content": "(v<sub>0</sub> + A)cos$$\\omega $$t sin$$\\omega $$<sub>0</sub> t"}, {"identifier": "C", "content": "v<sub>0</sub> sin[$$\\omega $$<sub>0</sub> (1+ 0.01Asin$$\\omega $$t)t]"}, {"identifie... | ["D"] | null | $$A = \left( {{v_0} + A\cos \,\omega t} \right)\sin {\omega _o}t$$<br><br>
$$ \Rightarrow {v_0}\sin \left( {{\omega _o}t} \right) + {A \over 2}\left[ {\sin \left( {{\omega _o} - \omega } \right)t + \sin \left( {{\omega _o} + \omega } \right)t} \right]$$ | mcq | jee-main-2019-online-9th-april-morning-slot | 9,626 |
bmWBmlwB381aUWfKNg3rsa0w2w9jx7efdda | physics | communication-systems | modulation-&-demodulation | In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin(20000 $$\pi $$t) while modulating signal
is given by, m(t) = 2 sin (2000 $$\pi $$t). The values of modulation index and lower side band frequency are : | [{"identifier": "A", "content": "0.3 and 9 kHz"}, {"identifier": "B", "content": "0.5 and 10 kHz"}, {"identifier": "C", "content": "0.4 and 10 kHz"}, {"identifier": "D", "content": "0.5 and 9 kHz"}] | ["D"] | null | C(t) = 4sin(20000 $$\pi $$t), A<sub>c</sub> = 4
<br><br>m(t) = 2sin(2000$$\pi $$t), A<sub>m</sub> = 2
<br><br>Modulation index (m) = $${{{A_m}} \over {{A_c}}}$$ = $${2 \over 4}$$ = 0.5
<br><br>Lower side band frequency = f<sub>c</sub> - f<sub>m</sub>
<br><br>= $${{20000\pi } \over {2\pi }}$$ - $${{2000\pi } \over {2\pi... | mcq | jee-main-2019-online-12th-april-evening-slot | 9,628 |
zZVQAsn00aRVZjQ74Hjgy2xukeuu8boy | physics | communication-systems | modulation-&-demodulation | An amplitude modulated wave is represented
by the expression<br/> v<sub>m</sub> = 5(1 + 0.6 cos 6280t)
sin(211 × 10<sup>4</sup> t) volts
<br/>The minimum and maximum amplitudes of the
amplitude modulated wave are, respectively | [{"identifier": "A", "content": "$${3 \\over 2}$$ V, 5 V"}, {"identifier": "B", "content": "$${5 \\over 2}$$ V, 8 V"}, {"identifier": "C", "content": "5 V, 8 V"}, {"identifier": "D", "content": "3 V, 5 V"}] | ["B"] | null | v<sub>m</sub> = 5(1 + 0.6 cos 6280t)
sin(211 × 10<sup>4</sup> t)
<br><br>$$ \Rightarrow $$ v<sub>m</sub> = (5 + 3 cos 6280t)
sin(211 × 10<sup>4</sup> t)
<br><br>maximum Amp. = 5 + 3 = 8 V
<br><br>minimum Amp. = 5 – 3 = 2 V
<br><br>From the given option nearest value of minimum Amplitude = $${5 \over 2}$$ V | mcq | jee-main-2020-online-2nd-september-morning-slot | 9,629 |
LARAAkNAf3utP5wOaR1klril76f | physics | communication-systems | modulation-&-demodulation | An audio signal v<sub>m</sub> = 20 sin 2$$\pi$$(1500t) amplitude modulates a carrier v<sub>c</sub> = 80 sin 2$$\pi$$(100,000t).<br/><br/>The value of percent modulation is _________. | [] | null | 25 | Given, audio signal,<br/><br/>V<sub>m</sub> = 20 sin 2$$\pi$$(1500t) .... (i)<br/><br/>Carrier signal, V<sub>c</sub> = 80 sin 2$$\pi$$(100000t) .... (ii)<br/><br/>We know that, modulation index,<br/><br/>$${m_f} = {{{A_m}} \over {{A_c}}}$$<br/><br/>From Eqs. (i) and (ii), we get<br/><br/>A<sub>m</sub> = 20, A<sub>c</su... | integer | jee-main-2021-online-24th-february-morning-slot | 9,630 |
oqikjGn7NvOIuvVumf1klry50o4 | physics | communication-systems | modulation-&-demodulation | Given below are two statements :<br/><br/>Statement I : A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.<br/><br/>Statement II : The side band frequencies are 1002 kHz and 998 kHz.<br/><br/>In the light of the above statements, choose the correct... | [{"identifier": "A", "content": "Statement I is false but Statement II is true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Statement I is true but Statement II is false"}, {"identifier": "D", "content": "Both Statement I and Statement II are true"}] | ["D"] | null | Side band = (f<sub>c</sub> $$-$$ f<sub>m</sub>) to (f<sub>c</sub> + f<sub>m</sub>)<br><br>= (1000 $$-$$ 2) KHz to (1000 + 2) Khz<br><br>= 998 KHz to 1002 KHz<br><br>Band width = 2f<sub>m</sub><br><br>= 2 $$\times$$ 2 KHz<br><br>= 4 KHz<br><br>Both statements are true. | mcq | jee-main-2021-online-25th-february-morning-slot | 9,631 |
lNRWIpPUaTVtUPwsqc1klunyd3i | physics | communication-systems | modulation-&-demodulation | If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are _________. | [] | null | 9 | B. W. (Bandwidth) = 2 $$\times$$ maximum frequency at modulating signal<br><br>= 2 $$\times$$ 5kHz<br><br>= 10 kHz<br><br>$$ \therefore $$ No of stations accommodate<br><br>= $${{90} \over {10}}$$ = 9 | integer | jee-main-2021-online-26th-february-evening-slot | 9,634 |
iLEYTwRISqPWd0Iw6y1kmk825wo | physics | communication-systems | modulation-&-demodulation | A carrier signal C(t) = 25 sin(2.512 $$\times$$ 10<sup>10</sup>t) is amplitude modulated by a message signal m(t) = 5 sin(1.57 $$\times$$ 10<sup>8</sup>t) and transmitted through an antenna. What will be the bandwidth of the modulated signal? | [{"identifier": "A", "content": "50 MHz"}, {"identifier": "B", "content": "8 GHz"}, {"identifier": "C", "content": "1987.5 MHz"}, {"identifier": "D", "content": "2.01 GHz"}] | ["A"] | null | $$\beta = 2{f_{m(t)}}$$<br><br>$$ \Rightarrow $$ $$\beta = 2 \times {{1.57 \times {{10}^8}} \over {2\pi }}$$<br><br>$$ \Rightarrow $$ $$\beta = 50$$ MHz | mcq | jee-main-2021-online-17th-march-evening-shift | 9,635 |
1krpq4t8e | physics | communication-systems | modulation-&-demodulation | A carrier wave V<sub>c</sub>(t) = 160 sin(2$$\pi$$ $$\times$$ 10<sup>6</sup>t) volts is made to vary between V<sub>max</sub> = 200 V and V<sub>min</sub> = 120 V by a message <br/>signal V<sub>m</sub>(t) = A<sub>m</sub> sin(2$$\pi$$ $$\times$$ 10<sup>3</sup>t) volts. The peak voltage A<sub>m</sub> of the modulating sign... | [] | null | 40 | Given, V<sub>C</sub>(t) = 160 sin(2$$\pi$$ $$\times$$ 10<sup>6</sup>t) V<br/><br/>V<sub>max</sub> = 200 V, V<sub>min</sub> = 120 V and V<sub>m</sub>(t) = A<sub>m</sub> sin(2$$\pi$$ $$\times$$ 10<sup>3</sup>t) V.<br/><br/>Since, maximum amplitude,<br/><br/>A<sub>max</sub> = A<sub>m</sub> + A<sub>c</sub><br/><br/>$$\Righ... | integer | jee-main-2021-online-20th-july-morning-shift | 9,636 |
1krum7riw | physics | communication-systems | modulation-&-demodulation | In amplitude modulation, the message signal<br/><br/>V<sub>m</sub>(t) = 10 sin (2$$\pi$$ $$\times$$ 10<sup>5</sup>t) volts and <br/><br/>Carrier signal<br/><br/>V<sub>C</sub>(t) = 20 sin(2$$\pi$$ $$\times$$ 10<sup>7</sup> t) volts<br/><br/>The modulated signal now contains the message signal with lower side band and up... | [{"identifier": "A", "content": "200 kHz"}, {"identifier": "B", "content": "50 kHz"}, {"identifier": "C", "content": "100 kHz "}, {"identifier": "D", "content": "0"}] | ["A"] | null | Bandwidth = 2 $$\times$$ f<sub>m</sub><br><br>= 2 $$\times$$ 10<sup>5</sup> HZ = 200 KHZ | mcq | jee-main-2021-online-25th-july-morning-shift | 9,637 |
1krwc9bvp | physics | communication-systems | modulation-&-demodulation | A message signal of frequency 20 kHz and peak voltage of 20 volt is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 volt. The modulation index will be : | [] | null | 1 | Modulation index<br><br>$$\mu = {{{A_m}} \over {{A_c}}} = {{20} \over {20}} = 1$$ | integer | jee-main-2021-online-25th-july-evening-shift | 9,638 |
1kryz2frr | physics | communication-systems | modulation-&-demodulation | The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5V amplitude are $${a \over {10}}V$$ and $${b \over {10}}V$$ respectively. Then the value of $${a \over b}$$ is ____________. | [] | null | 1 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264406/exam_images/fs8jz4onnrsunnejjp2j.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Physics - Communication Systems Question 56 English Explanation"><br><br>$... | integer | jee-main-2021-online-27th-july-morning-shift | 9,639 |
1ktagjv6r | physics | communication-systems | modulation-&-demodulation | An amplitude modulated wave is represented by <br/>C<sub>m</sub>(t) = 10(1 + 0.2 cos 12560t) sin(111 $$\times$$ 10<sup>4</sup>t) volts. The modulating frequency in kHz will be ................. . | [] | null | 2 | W<sub>m</sub> = 12560 = 2$$\pi$$f<sub>m</sub><br><br>$${f_m} = {{12560} \over {2\pi }}$$<br><br>= 2000 Hz | integer | jee-main-2021-online-26th-august-morning-shift | 9,641 |
1ktjq74ut | physics | communication-systems | modulation-&-demodulation | A bandwidth of 6 MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6 kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be __________. | [] | null | 500 | Signal bandwidth = 2 fm = 12 kHz<br><br>$$\therefore$$ N = $${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$$ | integer | jee-main-2021-online-31st-august-evening-shift | 9,642 |
1ktmwsh7o | physics | communication-systems | modulation-&-demodulation | A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : x, then value of x is ____________. | [] | null | 200 | Given, the amplitude of the carrier wave, A<sub>c</sub> = 250 V<br/><br/>The amplitude of the message wave, A<sub>m</sub> = 150 V<br/><br/>We know that,<br/><br/>The maximum amplitude, A<sub>max</sub> = A<sub>c</sub> + A<sub>m</sub><br/><br/>Substituting the values in the above equation, we get<br/><br/>A<sub>max</sub>... | integer | jee-main-2021-online-1st-september-evening-shift | 9,643 |
1l547tivn | physics | communication-systems | modulation-&-demodulation | <p>Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission :</p> | [{"identifier": "A", "content": "375 $$\\times$$ 10<sup>7</sup>"}, {"identifier": "B", "content": "75 $$\\times$$ 10<sup>7</sup>"}, {"identifier": "C", "content": "375 $$\\times$$ 10<sup>8</sup>"}, {"identifier": "D", "content": "75 $$\\times$$ 10<sup>9</sup>"}] | ["B"] | null | <p>$$v = f\lambda $$</p>
<p>$$ \Rightarrow f = {v \over \lambda } = {{3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$$ Hz $$ = 3 \times {10^{14}}$$ Hz</p>
<p>$$\Rightarrow$$ Channels $$ = {{{2 \over {100}} \times 3 \times {{10}^{14}}} \over {8 \times {{10}^3}}} = 75 \times {10^7}$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift | 9,644 |
1l55lokh4 | physics | communication-systems | modulation-&-demodulation | <p>Amplitude modulated wave is represented by $${V_{AM}} = 10\,[1 + 0.4\cos (2\pi \times {10^4}t)]\cos (2\pi \times {10^7}t)$$. The total bandwidth of the amplitude modulated wave is :</p> | [{"identifier": "A", "content": "10 kHz"}, {"identifier": "B", "content": "20 MHz"}, {"identifier": "C", "content": "20 kHz"}, {"identifier": "D", "content": "10 MHz"}] | ["C"] | null | <p>Bandwidth = 2 $$\times$$ f<sub>m</sub></p>
<p>= 2 $$\times$$ 10<sup>4</sup> Hz = 20 kHz</p> | mcq | jee-main-2022-online-28th-june-evening-shift | 9,645 |
1l58ibixt | physics | communication-systems | modulation-&-demodulation | <p>A sinusoidal wave y(t) = 40sin(10 $$\times$$ 10<sup>6</sup> $$\pi$$t) is amplitude modulated by another sinusoidal wave x(t) = 20sin (1000 $$\pi$$t). The amplitude of minimum frequency component of modulated signal is :</p> | [{"identifier": "A", "content": "0.5"}, {"identifier": "B", "content": "0.25"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "10"}] | ["D"] | null | <p>Modulate signal $$s(t) \equiv [1 + 20\sin (1000\pi t)]\sin ({10^7}\pi t)$$</p>
<p>$$ \equiv \sin ({10^7}\pi t) + 10\cos ({10^7}\pi t - {10^3}\pi t) + 10\cos ({10^7}\pi t + {10^3}\pi t)$$</p>
<p>$$\Rightarrow$$ Required amplitude = 10</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 9,647 |
1l5c457ob | physics | communication-systems | modulation-&-demodulation | <p>A baseband signal of 3.5 MHz frequency is modulated with a carrier signal of 3.5 GHz frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?</p> | [{"identifier": "A", "content": "42.8 m"}, {"identifier": "B", "content": "42.8 mm"}, {"identifier": "C", "content": "21.4 mm"}, {"identifier": "D", "content": "21.4 m"}] | ["C"] | null | <p>$${v _c} = 3.5 \times {10^9}$$ Hz</p>
<p>$$\therefore$$ $$\lambda = {c \over {{v _c}}} = {{3 \times {{10}^8}} \over {3.5 \times {{10}^9}}}$$</p>
<p>$$\therefore$$ Size of antenna $$ = {\lambda \over 4}$$</p>
<p>$$ = {{8.57 \times {{10}^{ - 2}}} \over 4}$$</p>
<p>$$ = 21.4$$ mm</p> | mcq | jee-main-2022-online-24th-june-morning-shift | 9,649 |
1l5w379rv | physics | communication-systems | modulation-&-demodulation | <p>A speech signal given by 11 sin(2200 $$\pi$$t) V is used for amplitude modulation with a carrier signal given by 44 sin(6600 $$\pi$$t) V. The minimum amplitude of modulated wave will be :</p> | [{"identifier": "A", "content": "33 V"}, {"identifier": "B", "content": "55 V"}, {"identifier": "C", "content": "8.25 V"}, {"identifier": "D", "content": "13.75 V"}] | ["A"] | null | $\mathrm{A}_{\min }=\mathrm{A}_{\mathrm{C}}-\mathrm{A}_{\mathrm{m}}$
<br/><br/>
$$
=(44-11) \text { volt }=33 \text { volt }
$$ | mcq | jee-main-2022-online-30th-june-morning-shift | 9,650 |
1l6i0vaz7 | physics | communication-systems | modulation-&-demodulation | <p>The maximum and minimum voltage of an amplitude modulated signal are $$60 \mathrm{~V}$$ and $$20 \mathrm{~V}$$ respectively. The percentage modulation index will be :</p> | [{"identifier": "A", "content": "0.5%"}, {"identifier": "B", "content": "50%"}, {"identifier": "C", "content": "2%"}, {"identifier": "D", "content": "30%"}] | ["B"] | null | <p>Percentage modulation</p>
<p>$$\mu = {{{V_{\max }} - {V_{\min }}} \over {{V_{\max }} + {V_{\min }}}} \times 100$$</p>
<p>$$ = {{60 - 20} \over {60 + 20}} \times 100$$</p>
<p>$$ = 50\% $$</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 9,653 |
1l6knxnze | physics | communication-systems | modulation-&-demodulation | <p>A square wave of the modulating signal is shown in the figure. The carrier wave is given by $$C(t)=5 \sin (8 \pi t)$$ Volt. The modulation index is :</p>
<p><img src="data:image/png;base64,UklGRoQIAABXRUJQVlA4IHgIAADQjACdASoAA28BP4HA3GQ2MbumodPZu3AwCWlu4W2jWmNwvV6Xje3cfir4LVU15TzOKKQG5Wvmk2vWp/JhUu+cUUgN/ZABv7IAN/ZA... | [{"identifier": "A", "content": "0.2"}, {"identifier": "B", "content": "0.1"}, {"identifier": "C", "content": "0.3"}, {"identifier": "D", "content": "0.4"}] | ["A"] | null | <p>$$\mu = {{{A_m}} \over {{A_c}}}$$</p>
<p>$$ = {1 \over 5} = 0.2$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift | 9,654 |
1l6mb111e | physics | communication-systems | modulation-&-demodulation | <p>In the case of amplitude modulation to avoid distortion the modulation index $$(\mu)$$ should be :</p> | [{"identifier": "A", "content": "$$\\mu \\leq 1$$"}, {"identifier": "B", "content": "$$\\mu \\geq 1$$"}, {"identifier": "C", "content": "$$\\mu = 2$$"}, {"identifier": "D", "content": "$$\\mu = 0$$"}] | ["A"] | null | <p>For effective modulation,</p>
<p>$$\mu$$ $$\le$$ 1</p> | mcq | jee-main-2022-online-28th-july-morning-shift | 9,655 |
1l6nt20q5 | physics | communication-systems | modulation-&-demodulation | <p>A FM Broadcast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is :</p> | [{"identifier": "A", "content": "220 kHz"}, {"identifier": "B", "content": "180 kHz"}, {"identifier": "C", "content": "360 kHz"}, {"identifier": "D", "content": "440 khz"}] | ["D"] | null | <p>Bandwidth of FM wave $$ = 2(\Delta f + {f_m})$$</p>
<p>$${{\Delta f} \over {{f_m}}} = 10$$ (Given)</p>
<p>$$\Delta f = {f_m}(10) = 20 \times 10 = 200$$ kHz</p>
<p>$$BW = 2(200 + 20)$$ kHz</p>
<p>$$ = 440$$ kHz</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 9,656 |
1l6p5hrin | physics | communication-systems | modulation-&-demodulation | <p>Find the modulation index of an AM wave having $$8 \mathrm{~V}$$ variation where maximum amplitude of the AM wave is $$9 \mathrm{~V}$$.</p> | [{"identifier": "A", "content": "0.8"}, {"identifier": "B", "content": "0.5"}, {"identifier": "C", "content": "0.2"}, {"identifier": "D", "content": "0.1"}] | ["A"] | null | <p>In an Amplitude Modulated (AM) signal, the amplitude of the carrier signal is varied in accordance with the information being sent. The extent of this variation is given by the modulation index, often denoted by $\mu$. </p>
<p>In the problem statement, we are given a peak-to-peak variation in the signal of 8V. This ... | mcq | jee-main-2022-online-29th-july-morning-shift | 9,657 |
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