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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
A disc of mass 5 kg and radius 50 cm rolls on the ground at the rate of `10 ms^(-1)`. Calculate the K.E. of the disc. |
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Answer» Given, mass of the disc, `M = 5 kg` Radius of the disc, `R=50 "cm"=(1)/(2)m` Linear velocity of the disc, `v=10 ms^(-1)` `because v = R omega rArr 10 = (1)/(2) omega` `rArr omega = 10 xx 2 =20 "rad s"^(-1)` Also, moment of inertia pf disc about an axismk through its centre, `I=(1)/(2)MR^(2)` `rArr I=(1)/(2)xx5xx((1)/(2))^(2)=(5)/(8)"kg-m"^(2)` `because` Kinetic enegry of translation `= (1)/(2) mv^(2)` `=(1)/(2) xx5xx(10)^(2)=250 J` `because` Kinetic energy of rotation `= (1)/(2) I omega^(2)` `= (1)/(2)xx(5)/(8) xx(20)^(2)=125 J` `:.` Total kinetic energy `- (250+125) J 375 J`. |
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| 52. |
Give reasons for each of the following : The length of day and night is not equal at all places on the earth. |
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Answer» Due to the tilted axis of the’earth and the migration belt of the sun between Tropics of Cancer and. Capricorn, the length of day and night differ from place to place and region to region. |
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| 53. |
Account for the unequal length of day and night. |
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Answer» The length of day and night varies throughout the year. Sometimes the days are longer and nights are shorter and vice versa. It is due to two reasons : (a) The inclination of the axis. (b) The revolution of the Earth. Due to inclined axis, one hemisphere leans towards the Sun for the six months ; the other Hemisphere leans towards the Sun for the next six months. In summer, on 21st June, days are longer in northern Hemisphere and the nights are shorter. In winter, the conditions are reversed, the days are shorter and nights are longer in the Northern Hemisphere. If the axes were vertical, there would have been equal days and nights everywhere |
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| 54. |
Mention one effect of seasons in low and high latitudes. |
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Answer» The effect of seasons in low and high latitudes is distinct through various phases of different seasons. Low latitude areas get ample temperature and rainfall while the high latitude areas are cold and receive less rainfall. Tropical regions are always warm with heavy rainfall season, while the temperate and polar regions are cold with scanty rainfall, while the polar regions are always covered with perpetual snow due to the temperature below 0°C. |
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| 55. |
What are the effects of the inclination of the axis ? |
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Answer» (a) There is apparent movement of the Sun from the equator towards tropics. (b) It causes opposite seasons in the two Hemisphere. (c) It results in the variation of length of day and night. (d) When North Pole is tilted towards the Sun, the South Pole turns away from the Sun. So the two Hemispheres are alternately exposed to the Sun rays. |
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| 56. |
What are the results of the difference in the Earth’s speed of rotation at various latitudes ? |
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Answer» At equator, the Earth’s speed of rotation is about 1600 km per hour. The speed, goes on decreasing poleward. At 60° latitude, it is 800 km per hour. This difference affects the general circulation of the atmosphere. Winds are deflected to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. Similarly, the ocean currents are also deflected. This is known as Ferral’s Law. Due to this, the Earth is flattened at the poles and bulging at the equator. |
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| 57. |
Distinguish between the following pairs : (a) Summer Solstice and Winter Solstice. (b) Solstice and Equinoxes |
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Answer» (a) Summer Solstice : 1. The Earth is in this position on its orbit on 21st June. 2. The Sun is overhead on Tropic of Cancer (23 1/2° N). 3. Days are longer than nights in the Northern Hemisphere. 4. There is summer season in the Northern and winter season in the Southern Hemisphere. Winter Solstice : 1. The Earth is in this position on its orbit on 22nd December. 2. The Sun is overhead on Tropic of Capricorn (23 1/2° S). 3. Days are shorter than night in the Northern Hemisphere. 4. There is winter season in the Northern and summer season in the Southern Hemisphere. (b) Solstices : 1. The positions of the Earth on 21st June and 22nd December are known as Solstices. 2. The Sun is overhead at tropics. 3. Days and nights are unequal. 4. The term Solstice mean “Sun standing ‘still’. 5. There is summer or winter season in Northern Hemisphere. Equinoxes : 1. The positions of the Earth on 23rd September and 21st March are known as Equinoxes. 2. The Sun is overhead at Equator. 3. Days and nights are equal. 4. The term Equinox means ‘Equal nights’. 5. There is autumn or spring season in Northern Hemisphere |
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| 58. |
Two thin uniform rings made of same material and of radii `R` and `4 R` are joined as shown. The mass of smaller ring is `m`. Find the `M.I.` about an axis passing through the center of mass of system of rings and perpendicular to the plane. |
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Answer» Here, `m_(1)=m,m_(2)=4m` Location of the centre of mass from O `X_(CM)=(mxxR+4m(2R+4R))/(m+4m)=5R` A is CM of system of rings `I_(A)=I_(1)+I_(2)={mR^(2)+m(5R-R^(2))}+{4m(4R)^(2)+4m(6R-5R)^(2)}` `=mR^(2)[{1+16}+{64+4}]` `I_(A)=85 mR^(2)`. |
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| 59. |
The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height `h` from rest without slipping will be.A. `sqrt((10 gh)/(7))`B. `sqrt((6gh)/(5))`C. `sqrt((4 gh)/(5))`D. `sqrt(2gh)` |
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Answer» Correct Answer - B In pure rolling, mechanical energy remains conserved. Therefore, at bottomost post, total kinetic energy will be `mgh` Ratio of rotational to translational kinetic energy will be `(2)/(3)` `:. K_(T)=(3)/(5)(mgh)rArr(3)/(5)mgh=(1)/(2)mv^(2)` `:.` The speed of a uniform spherical shell `v=sqrt((6gh)/(5))`. |
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| 60. |
A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of `15^@` in time `t` sec. Then how much it rotates in the next `2t` secA. `90^(@)`B. `120^(@)`C. `30^(@)`D. `45^(@)` |
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Answer» Correct Answer - B If angular acceleration is constant, we have `theta=omega_(0)t+(1)/(2)alphat^(2)` Given, `theta=15^(@), omega_(0)=0` For the 1st condition (time = t s) `15^(@)=(1)/(2) alphat^(2)`….(i) For the 2nd condition (time = 3t s) `theta_(1)=(1)/(2)alpha(3t)^(2)=(1)/(2)(alpha)9t^(2)`....(ii) So, the increase in angle through which it rotates in the next 2s, `Deltatheta=theta_(1)=(1)/(2)alphat^(2)` `=9xx(1)/(2)alphat^(2)-(1)/(2)alphat^(2)=8((1)/(2)alphat^(2))` `=8xx15^(@) =120^(@)` [from eq. (i)]. |
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| 61. |
The ratio of the time taken by a solid sphere and that taken by a disc of the same mass and radius to roll down a rough inclined plane from rest, from the same height isA. `15:14`B. `sqrt(15) : sqrt(14)`C. `14 : 15`D. `sqrt(14) : sqrt(15)` |
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Answer» Correct Answer - D In case of pure rolling, `a=(g sin theta)/(1+(I)/(mR^(2)))` `(I)/(mR^(2))=(2)/(5)` for sphere and `(I)/(MR^(2))=(1)/(2)` for a disc `:.` For a solid sphere, `a_(1)=(5)/(7)g sin theta` and for a hollow sphere, `a_(2)=(2)/(3)g sin theta` From `s=(1)/(2)at^(2),t=sqrt((2s)/(a))or (t_(1))/(t_(2))=sqrt((a_(2))/(a_(1)))=sqrt((2//3)/(5//7))=sqrt((14)/(15))`. |
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| 62. |
A car moves on a circular road. It describes equal angles about the centre in equal intervals of time. Which of the following statement about the velocity of the car is trueA. Magnitude of velocity is not constantB. Both magnitude and direction of velocity changeC. Velocity is directed towards the centre of the circleD. Magnitude of velocity is constant but direction changes. |
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Answer» Correct Answer - D As body covers equal angles in equal intervals of time, so its angular velocity and hence, magnitude of linear velocity is constant but its direction changes `( :. V = r omega)`. |
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| 63. |
A particle is moving in a circular orbit with constant speed. Select wrong alternate.A. Its linear momentum is conservedB. Its angular momentum is conservedC. It is moving with variable velocityD. It is moving with variable acceleration |
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Answer» Correct Answer - A Direction of linear velocity always keeps on changing. Hence, linear momentum is varying. |
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| 64. |
Choose the wrong statement.A. The centre of mass of a uniform circular ring is at its geometric centreB. Moment of inertia is scalar quantityC. Radius of gyration is a vector quantityD. Force in translational motion is analogous to torque in rotational motion |
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Answer» Correct Answer - B Moment of inertia is not a scalar quantity so the option (b) is wrong. |
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| 65. |
A sphere is rolling down a plane of inclination `theta` to the horizontal. The acceleration of its centre down the plane isA. `g sin theta`B. less than `g sin theta`C. greater than `g sin theta`D. zero |
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Answer» Correct Answer - B The acceleration of its centre down the plane `a=(g sin theta)/(1+(I)/(mR^(2)))or alt gsin theta` . |
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| 66. |
A small object of uniform density rolls up a curved surface with an initial velocity v. it reaches up to a maximum height of `(3v^2)/(4g) with respect to the initial position. The object isA. ringB. solid sphereC. hollow sphereD. disc |
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Answer» Correct Answer - D As, `v=sqrt((2gh)/(1+K^(2)//R^(2)))` Given, `h=(3v^(2))/(4g)rArrv^(2)=(2gh)/(1+(K^(2))/(R^(2)))=(2g 3v^(2))/(4g(1+(K^(2))/(R^(2))))=(6gv^(2))/(4g(1+(K^(2))/(R^(2))))` `1=(3)/(2(1+K^(2)//R^(2)))or1+(K^(2))/(R^(2))=(3)/(2)` or `(K^(2))/(R^(2))=(3)/(2)-1=(1)/(2)rArrK^(2)=(1)/(2)R^(2)` (Equation of disc) Hence, the object is disc. |
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| 67. |
A motor is rotating at a constant angular velocity of 500 rpm. Thw angular displacement per second isA. `(3)/(50 pi)` radB. `(3 pi)/(50)` radC. `(25 pi)/(3)` radD. `(50 pi)/(3)` rad |
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Answer» Correct Answer - D Angular displacement per second `theta=omegat=(500xx2pi)/(60)=(50pi)/(3)"rad"`. |
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| 68. |
The motor of an engine is rotating about its axis with an angular velocity of 100 rpm. It comes to rest in 15 s after being switched off. Assuming constant angular deceleration, calculate the number of revolution made by it before coming to rest.A. 12.5B. 40C. 32.6D. 15.6 |
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Answer» Correct Answer - A From the equation, `0 = omega_(0) - alpha t` `:. alpha=(omega_(0))/(t)=((100xx2pi)//60)/(15)=0.7 "rad s"^(-2)` Now angle rotated before coming to rest, `theta = (omega_(0)^(2))/(2 alpha)` or `theta =((100xx2pi//60)^(2))/(2xx0.7)=78.33` rad or number rotations, `n=(theta)/(2pi)=12.5`. |
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| 69. |
A wheel is rotating at the rate of `33 "rev min"^(-1)`. If it comes to stop in 20 s. Then, the angular retardation will beA. `pi rad s^(-2)`B. `11 pi rads^(-2)`C. `(pi)/(200) rads^(-2)`D. `(11 pi)/(200) rads^(-2)` |
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Answer» Correct Answer - D Angular retardation, `alpha=(omega_(0))/(t)" "(because 0=omega_(0)-alphat)` `=(2pixx33//60)/(20)=(11pi)/(200) "rad s"^(-2)`. |
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| 70. |
A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 min. The angular retardation in `rad s^(-2)` isA. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(8)` |
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Answer» Correct Answer - A Angular velocity, `omega_(0)=(900xx2pi)/(60)"rad s"^(-1)=30 pi "rad s"^(-1)` Now, `0=omega_(0)-alphat` or Angular retardation, `alpha=(omega_(0))/(t)=(30pi)/(60)"rad s"^(-2) =(pi)/(2) "rad s"^(-2)`. |
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| 71. |
A diver in a swimming pool bends his head before diving, because itA. decreases his moment of inertiaB. decreases his angular velocityC. increases his moment of inertiaD. decreases his linear velocity |
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Answer» Correct Answer - A On bending the head before diving in swimming pool, the moment of inertia of a person decreases. |
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| 72. |
A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]A. 5 cmB. 0.5 cmC. `sqrt(10)` cmD. `sqrt(5)` m |
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Answer» Correct Answer - D The relation between linear velocity `(v)` and angular velocity `(omega)` is `v=r omegarArr r=(v)/(omega)` Total kinetic energy `= (1)/(2)mv^(2)+(1)/(2)Iomega^(2)` `=(1)/(2)mv^(2)+(1)/(2)xx(1)/(2)mr^(2)omega^(2)=(1)/(2)mv^(2)+(1)/(4)mv^(2)=(3)/(4)mv^(2)` `(3)/(4)mv^(2)=mg.3rArrv=2 sqrt(10)` `because omegar=vrArrr=(v)/(omega)=(2sqrt(10))/(2 sqrt(2))=sqrt(5)` m. |
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| 73. |
Assertion : A sphere is placed in pure rolling condition over a rough inclined surface.Then, force of friction will act in downward direction Reason : Angular acceleration (actually retardation) due to friction is anti-clockwise.A. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is fasleD. If Assertion is false but Reason is true |
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Answer» Correct Answer - D Linear acceleration due to `mg sin theta` is downward. Hence for pure rolling to continue, force of friction should act upwards so that angular acceleration is anti-clockwise. |
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| 74. |
At any instant, a rolling body may be considered to be in pure rotation about an axis through the point of contact. This axis is translating forward with speedA. equal to centre of massB. zeroC. twice of centre of massD. None of these |
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Answer» Correct Answer - A Since, in this case instantaneous axis of rotation is always below the centre of mass. This is possible only when point of contact moves with a velocity equal to centre of mass. |
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| 75. |
The ratio of the radii of gyration of a hollow sphere and a solid sphere of the same radii about aA. `sqrt((7)/(3))`B. `(5)/(sqrt(21))`C. `sqrt((21)/(5))`D. `(25)/(9)` |
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Answer» Correct Answer - B Hollow sphere, `K_(H)=sqrt((I)/(m))=sqrt(((2)/(3)mR^(2)+mR^(2))/(m))=sqrt((5)/(3))R` Solid sphere, `K_(S)=sqrt((I)/(m))=sqrt(((2)/(5)mR^(2)+mR^(2))/(m))=sqrt((7)/(5))R` `:. (K_(H))/(K_(S))=sqrt((25)/(21))=(5)/(sqrt(21))`. |
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| 76. |
Find the torque of a force `F=(hat(i)2 +hat(j)-3hat(k))N` about a point O. The position vector of point of application of force about O is `r = (2 hat(i) + 3 hat(j) - hat(k))m`. |
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Answer» Torque, `tau=rxxF=|{:(hat(i),hat(j),hat(k)),(2,3,-1),(1,2,-3):}|` `tau=hat(i) (-9+2)+hat(j)(-1+6)+hat(k)(4-3)` or `tau=(-7hat(i)+5 hat(j)+hat(k))"Nm"`. |
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| 77. |
A force of `-4Fhat(K)` acts O, the origin of the coordinate system. The torque about the point `(1,-1)` is |
| Answer» Torque, `tau=rxxF=[(1-0)hat(i)+(-1-0)hat(j)+(0-0)hat(k)]xx(-10)hat(k)=10(hat(i)-hat(j))xx(-hat(k))=10(hat(i)+hat(j))Nm` | |
| 78. |
A body rotating with uniform angular acceleration covers `100 pi` (radian) in the first 5 s after the start. Its angular speed at the end of 5 s (in rad/s) isA. `40 pi`B. `30 pi`C. `20 pi`D. `10 pi` |
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Answer» Correct Answer - C `theta=100 pi,t-=5 s, omega=?` Angular speed,`omega=(theta)/(t)=(100pi)/(5)=20pi`. |
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