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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.
| 151. |
Which of the following is the altitude-time graph for a projectile thrown horizontally from the top of the towerA. B. C. D. |
| Answer» Correct Answer - D | |
| 152. |
A ball is projected from the top of a tower at an angle `60^(@)` with the vertical. What happens to the vertical component of its velocity?A. Increases continuouslyB. Decreases continuouslyC. Remain unchangedD. Frist decreases and then increases |
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Answer» Correct Answer - a As acceleration due to gravity acts against the motion up to the highest point, hence vertical component of the velocity first decreases. But during downwards motion, acceleration due to gravity acts in the direction of motion, hence vertical component of velocity then starts increasing. |
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| 153. |
A particle covers equal distance aroung a circular path, in equal intervals of time. Which of the following quntities connected with the motion of the particle remains constant with time?A. displacementB. velocityC. speedD. acceleration |
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Answer» Correct Answer - b Displacement, velocity and acceleration change continuously with respect to time because of change in direction. |
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| 154. |
Find the angle through which a cyclist bends when he covers a circular path `34.3 m` long in `sqrt (22)` sec . Given `g = 9.8 ms^(-2)` .A. `45^(@)`B. `40^(@)`C. `42^(@)`D. `48^(@)` |
| Answer» Correct Answer - A | |
| 155. |
If a particle covers half the circle of radius R with constant speed thenA. Momentum change is mvrB. Change in K.E is `1//2 mv`C. Change in K.E is mvD. Change in K.E is zero |
| Answer» Correct Answer - D | |
| 156. |
A ball is thrown upwards at an angle of `60^(@)` to the horizontal. It falls on the ground at a distance of 90 m . If the ball is thrown with the same initial velocity at an angle `30^(@)`, it will fall on the ground at a distance ofA. `30m`B. `60 m`C. `90 m`D. `120 m` |
| Answer» Correct Answer - C | |
| 157. |
A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speedA. `100m`B. `150 m`C. `200 m`D. `250 m` |
| Answer» Correct Answer - C | |
| 158. |
For a given velocity, a projectile has the same range `R` for two angles of rpojection if `t_(1)` and `t_(2)` are the times of flight in the two cases thenA. `t_(1)t_(2) prop R^(2)`B. `t_(1)t_(2) prop R`C. `t_(1)t_(2) prop (1)/(2)`D. `t_(1)t_(2) prop (1)/(R^(2))` |
| Answer» Correct Answer - B | |
| 159. |
Two identical balls are projected, one vertically up and the other at an angle of `30^(@)` to the horizontal, with same initial speed. The potential energy at the highest point is in the ratio:A. `4:3`B. `3:4`C. `4:1`D. `1:4` |
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Answer» Correct Answer - c Using equation of motion, height reached by first body, `h_(1)=(u^(2))/(2g)` height reached by second body, `h_(2)=(u^(2)sin^(3) 30^(@))/(2g)` `=1/4xx(u^(2))/(2g)` `:. h_(1):h_(2)=4:1` As `PE prop height` `:. (PE)_(1):(PE)_(2)=4:1` |
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| 160. |
For a projectile, the ratio of maximum height reached to the square of flight time is `(g = 10 ms^(-2))`A. `5:4`B. `5:2`C. `5:1`D. `10:1` |
| Answer» Correct Answer - A | |
| 161. |
When a projectile is fired at an angle `theta` w.r.t horizontal with velocity u, then its vertical component:A. remains sameB. goes on increasing with heightC. goes on decreasing with heightD. first increase then decreases with height |
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Answer» Correct Answer - c The vertical component goes on decreasing and eventually becomes zero. |
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| 162. |
Neglecting the air resistance, the time of flight of a projectile is determined byA. `U_("vertical")`B. `U_("horizontal")`C. `U = U_("vertical")^(2) +U_("horizontal")^(2)`D. `U = U(U_("vertical")^(2) +U_("horizontal"))^(1//2)` |
| Answer» Correct Answer - A | |
| 163. |
A projectile is fired with initial momentum p at an angle `45^(@)` from a point P as shown in figure. Neglecting air resistance, the magnitude of change in momentum between leaving P and arriving at Q is: A. `p//2`B. `psqrt(2)`C. `p`D. `2p` |
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Answer» Correct Answer - b Initial momentum =p, Final momentum =-p Angle between momentum vectors `=270^(@)` |change in momentum| `=sqrt(p^(2)+(-p)^(2)+2p(-p)cos 270^(@))=sqrt(2)p` |
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| 164. |
A particle is moving in a circle of radius R in such a way that any insant the total acceleration makes an angle of `45^(@)` with radius. Initial speed of particle is `v_(0)`. The time taken to complete the first revolution is:A. `(2R)/(v_(0))(1-e^(-2pi))`B. `(R)/(v_(0))(1-e^(-2pi))`C. `R/(v_(0))`D. `(2R)/(v_(0))` |
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Answer» Correct Answer - b Total acceleration makes an anlge of `45^(@)` with radius i.e., tangential acceleration =radial acceleration. `R alpha=R omega^(2)` or `alpha=omega^(2)` or `(d omega)/(dt)=omega^(2)` or `(d omega)/(omega^(2))=dt` or `int_(omega_(0))^(omega) (domega)/(omega^(2))=int_(0)^(t)dt` `omega=(omega_(0))/(1-omega_(0)t)` or `(d theta)/(dt)=(omega_(0))/(1-omega_(0)t)` `int_(0)^(2pi)d theta=int_(0)^(t) (omega_(0)dt)/(1-omega_(0)t)` or `t= 1/(omega_(0))(1-e^(-2pi))` `=R/(v_(0))(1-e^(-2pi))` |
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| 165. |
A simple pendulum is oscillating without damiping, When the displacement of the bob is less than maximum, its acceleration vector `veca` is correctly show in:A. B. C. D. |
| Answer» Correct Answer - C | |
| 166. |
The greatest height to which a boy can throw a stone is (h). What will be the greatest distance on horizontal surface upto which the boy can throw the stone with the same speed ? Neglect the air friction.A. `(h)/(2)`B. hC. `2h`D. `3h` |
| Answer» Correct Answer - C | |
| 167. |
A person can throw a stone to a maximum height of h meter. The maximum distance to which he can throw the stone is:A. `h`B. `h//2`C. `2h`D. `3h` |
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Answer» Correct Answer - c Height H is given by `H=(u^(2) sin^(2) theta)/(2g)` When `theta =90^(@), H=H_(max)=(u^(2))/(2g)=h` Range is given by: `R=(u^(2) sin 2theta)/g` Now, `R_(max)=(u^(2))/g (When theta=45^(@))=2h` |
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| 168. |
A person can throw a stone to a maximum distance of h meter. The greatest height to which he can throw the stone is:A. `h`B. `h//2`C. `2h`D. `3h` |
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Answer» Correct Answer - b `R=(u^(2) sin 2theta)/g` When `2 theta=90^(@) then R=R_(max)=(u^(2))/g` Given `R_(max)=h, hence (u^(2))/g=h` Height H is given by : `H=(u^(2) sin^(2) theta)/(2g)` When `theta=90^(@), H=H_(max)=(u^(2))/(2g)=h/2` |
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| 169. |
Roads are banked on curves so thatA. The speeding vehicles may not fall outwardsB. The frictional force between the road and vehicle may be decreasedC. The wear and tear of tyres may be avoidedD. The weight of the vehicle may be decreased |
| Answer» Correct Answer - A | |
| 170. |
Assertion : A coin is placed on phonogram turn table. The motor is started, coin moves along the moving table. Reason : Rotating table is providing necessary centripetal force to the coin.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - D | |
| 171. |
A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will beA. DoubleB. HalfC. 4 timesD. `(1)/(4)` times |
| Answer» Correct Answer - C | |
| 172. |
The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is (assume `g = 10 ms^(-2)`)A. `40 ms^(-1)`B. `20ms^(-1)`C. `15 ms^(-1)`D. `10 ms^(-1)` |
| Answer» Correct Answer - D | |
| 173. |
A particle revolves round a circular path. The acceleration of the particle isA. Along the circumference of the circleB. Along the tangentC. Along the radiusD. Zero |
| Answer» Correct Answer - C | |
| 174. |
Certain neutron stars are believed to be rotating at about `1rev//sec` . If such a star has a radius of 20 km , the acceleration of an object on the equator of the star will beA. `20 xxx 10^(8)m//sec^(2)`B. `8 xx 10^(5) m//sec^(2)`C. `120 xx 10^(5) m//sec^(2)`D. `4 xx 10^(8) m//sec^(2)` |
| Answer» Correct Answer - B | |
| 175. |
A particle of mass M moves with constant speed along a circular path of radius r under the action of a force F. Its speed isA. `sqrt((r F)/(m))`B. `sqrt((F)/(r))`C. `sqrt(F m r)`D. `sqrt((F)/(m r))` |
| Answer» Correct Answer - A | |
| 176. |
In an atom for the electron to revolve around the nucleus, the necessary centripetal force is obtained from the following force exerted by the nucleus on the electroA. Nuclear forceB. Gravitational forceC. Magnetic forceD. Electrostatic force |
| Answer» Correct Answer - D | |
| 177. |
A body of mass `100g` is tied to one end of a `2m` long string. The other end of the string is at the centre of the horizontal circle. The maximum revolution in one minute is `200`. The maximum tensible strength of the string is approxA. 8.76 NB. 8.94 NC. 89.42 ND. 87.64 N |
| Answer» Correct Answer - D | |
| 178. |
A car is moving on a circular path and takes a turn. If `R_(1)` and `R_(2)` be the reactions on the inner and outer wheels, respectively, thenA. `R_(1) = R_(2)`B. `R_(1) lt R_(2)`C. `R_(1) gt R_(2)`D. `R_(1) ge R_(2)` |
| Answer» Correct Answer - B | |
| 179. |
Assertion : When an automobile while going too fast around a curve overturns, its inner wheels leave the ground first. Reason : For a safe turn the velocity of automobile should be less than the value of safe limit velocity.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - B | |
| 180. |
A car speeds up in a circular path. Which of the following figures illustrates the acceleration of the car?A. B. C. D. |
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Answer» Correct Answer - b Net acceleration: `a=sqrt(a_(c)^(2)+a_(t)^(2))` `=sqrt(((v^(2))/R)^(2)+a_(t)^(2))` As v increases, a also increases. So size of arrow should be increasing and angle between velocity and acceleration should be acute. |
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| 181. |
Assertion: When the body is dropped or thrown horizontally form the same height, it would reach the ground at the same time. Reason: Horizontal velocity has no effect on the vertical direction.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - a Both body will take same time to reach the earth because vertical downward component of velocity for both the bodies will be zero and time of decent `t=sqrt((2k)/g)`. Horizontal velocity has no effect on the vertical direction. |
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| 182. |
If R is the maximum horizontal range of a particle, then the greatest height attained by it is :A. `R`B. `2R`C. `R//2`D. `R//4` |
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Answer» Correct Answer - d `R=(u^(2))/g and H=(u^(2) sin^(2) theta)/(2g)` For maximum range, ` theta=45^(@)` `H=(u^(2) sin^(2) 45^(@))/(2g)=(u^(2))/(4g)=R/4` |
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| 183. |
During a projectile motion if the maximum height equal the horizontal range, then the angle of projection with the horizontal is :A. `tan^(-1)(1)`B. `tan^(-1)(2)`C. `tan^(-1)(3)`D. `tan^(-1)(4)` |
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Answer» Correct Answer - d `H=(u^(2) sin^(2) theta)/(2g) and R=(u^(2) sin 2 theta)/g` Since `H=R(u^(2) sin^(2) theta)/(2g) =(u^(2)xx2sin theta cos theta)/g` or `tan theta=4 or theta=tan^(-1)(4)` |
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| 184. |
A particle is projeted with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is :A. `(4v^(2))/(5g)`B. `(4g)/(5v^(2))`C. `(4v^(3))/(5g^(2))`D. `(4v)/(5g^(2))` |
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Answer» Correct Answer - a `H=(v^(2) sin^(2) theta)/(2g) and R=(v^(2) sin 2 theta)/g` Since `R=2H, (v^(2) sin 2 theta)/g=2xx(v^(2) sin^(2) theta)/(2g)` or `2 sin theta cos theta= sin^(2) theta or tan theta=2` `R=v^(2)xx2/gxxsin theta cos theta` `=(2v^(2))/g =2/(sqrtg)xx1/(sqrt5)=(4v^(2))/(5g)` |
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| 185. |
Which of the following sets of factors will affect the horizontal distance covered by an converted by an athlete in a long- jump event?A. Speed before he jumps and his weightB. The direction in which he leaps and the initial speedC. The force with which he pushes the ground and his speedD. The direction in which he leaps and the weight |
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Answer» Correct Answer - b `R=(u^(2)sin 2 theta)/g implies` R depends upon only R and `theta`. |
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| 186. |
A man runs along a horizontal road holding his umrella vertical in order to afford maximum protection form rain. The rain is actually.A. Falling verticalB. Comming from front of the manC. Coming from the back of the manD. Either of (a), (b) or (c) |
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Answer» Correct Answer - c The horizontal component of rain should have same direction and magnitude as the velocity of man. |
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| 187. |
A cyclist turns around a curve at 15 miles/hour. If he turns at double the speed, the tendency to overturn isA. DoubledB. QuadrupledC. HalvedD. unchanged |
| Answer» Correct Answer - B | |
| 188. |
What is the value of linear velocity, if `vec(omega) = 3hat(i)-4 hat(j) + hat(k)` and `vec(r) = 5hat(i)-6hat(j)+6hat(k)`A. `6hat(i)+2hat(j)-3hat(k)`B. `-18 hat(i)-13hat(j)+2hat(k)`C. `4hat(i)-13hat(j)+6hat(k)`D. `6hat(i)-2hat(j)+8hat(k)` |
| Answer» Correct Answer - B | |
| 189. |
An airplane moving horizontally with a speed of `18km//hr` drops a food packet while flying at a height of 500m. The horizontal range is:A. `180m`B. `980m`C. `500m`D. `670m` |
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Answer» Correct Answer - c `s=ut+1/2"gt"^(2)` `500=1/2xx10xxt^(2) or t=10 sec`. Horizontal range `x=(180xx5)/18xx10=500`. |
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| 190. |
An aeroplane moving horizontally with a speed of `720 km//h` drops a food pocket, while flying at a height of 396.9 m . the time taken by a food pocket to reach the ground and its horizontal range is (Take `g = 9.8 m//sec`)A. `3 sec` and `2000m`B. `5 sec` and `500m`C. `8 sec` and `1500 m`D. `9 sec` and `1800m` |
| Answer» Correct Answer - D | |
| 191. |
A proton of mass `1.6 xx 10^(-27)kg` goes round in a circular orbit of radius 0.10 m under a centripetal force of `4 xx 10^(-13) N`. then the frequency of revolution of the proton is aboutA. `0.08 xx 10^(8)` cycles per secB. `4 xx 10^(8)` cycles per secC. `8 xx 10^(8)` cycles per secD. `12 xx 10^(8)` cycles per sec |
| Answer» Correct Answer - A | |