InterviewSolution
Saved Bookmarks
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.
| 101. |
In case of uniform circular motion which of the following physical quantity do not remain constantA. SpeedB. MomentumC. Kinetic energyD. Mass |
| Answer» Correct Answer - B | |
| 102. |
In uniform circular motionA. Both the angular velocity and the angular momentum varyB. The angular velocity varies but the angular momentum remains constantC. Both the velocity and the angular momentum stay constantD. The angular momentum varies but the angular velocity remains constant |
| Answer» Correct Answer - C | |
| 103. |
When a particle moves in a uniform circular motion. It hasA. Radial velocity and radial accelerationB. Tangential velocity and radial accelerationC. Tangential velocity and tangential accelerationD. Radial velocity and tangential acceleration |
| Answer» Correct Answer - B | |
| 104. |
A scooter is going round a circular road of radius 100 m at a speed of `10 m//s` . The angular speed of the scooter will beA. `0.01 rad//s`B. `0.1 rad//s`C. `1 rad//s`D. `10 rad//s` |
| Answer» Correct Answer - B | |
| 105. |
The point from where a ball is projected is taken as the origin of the coordinate axes. The x and y components of its displacement are given by `x=6t and y=7t-5t^(2)`. What is the velocity of projection?A. `6ms^(-1)`B. `8ms^(-1)`C. `10ms^(-1)`D. `14ms^(-1)` |
|
Answer» Correct Answer - c `v_(x)=(dx)/(dt)=6 and v_(y)=(dy)/(dt)=8-10t` `=8-10xx0=0` `v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(6(2)+8^(2))=10ms^(-1)` |
|
| 106. |
The x and y coordinates of the particle at any time are `x=5t-2t^(2) and y=10t` respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2s is:A. `5m//s^(2)`B. `-4m//s^(2)`C. `-8m//s^(2)`D. `0` |
|
Answer» Correct Answer - b `v_(x)=5-4t, v_(y)=10` `a_(x)=-4, a_(y)=0` `veca=a_(x)hati+a_(y)hatj` `veca=-4hatim//s^(2)` |
|
| 107. |
A man can swim with a speed of `4kmh^(-1)` in still water. He crosses a river 1km wise that flows steadly at `3kmh^(-1)`. If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank?A. `500m`B. `600m`C. `750m`D. `850m` |
|
Answer» Correct Answer - c Time to cross river, `t=("Width of river")/("Speed of man")` `=(1km)/(4kmh^(-1))=1/4h` Distance moved along the river in time t `=V_(r)xxt=3km^(-1)xx1/4h=3/4km=750m` |
|
| 108. |
A `100 kg` car is moving with a maximum velocity of `9 m//s` across a circular track of radius 30 m . The maximum force of friction between the road and the car isA. `1000N`B. `706 N`C. `270 N`D. `200 N` |
| Answer» Correct Answer - C | |
| 109. |
Find the maximum velocity for skidding for a car moved on a circular track of radius 100 m . The coefficient of friction between the road and tyre is 0.2A. `0.14 m//s`B. `140 m//s`C. `1.4 km//s`D. `14 m//s` |
| Answer» Correct Answer - D | |
| 110. |
A gun is firing bullets with velocity `v_0` by rotating it through `360^@` in the horizontal plane. The maximum area covered by the bullets isA. `pi((u^(2))/g)^(2)`B. `pi((u^(2))/(2g))^(2)`C. `pi(u/g)^(2)`D. `pi(u/(2g))^(2)` |
|
Answer» Correct Answer - a `R=(u^(2)sin 2 theta)/g` Range R is maximum when ` sin 2 theta=1` `R_(max)=(u^(2))/g` Maximum area coverted `=pi(R_(max))^(2)=pi((u^(2))/g)^(2)` |
|
| 111. |
The maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increases in horizontal range?A. `5%`B. `10%`C. `15%`D. `20%` |
|
Answer» Correct Answer - a If h be the maximum height attained by the projectile then `h=(u^(2)sin^(2) theta)/(2g) or R=(u^(2)sin2 theta)/g` `R/h=(2 sin thetacos theta)/((sin^(2) theta)//2) = 4 coth theta` Therefore `(DeltaR)/R=(Deltah)/h` `:.` Percentage increases in R=percentage increases in `y_(m)=5%` |
|
| 112. |
Assertion: A uniform circular motion is an acceleration motion. Reason: Direction of acceleration is parallel to velocity vector.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. |
|
Answer» Correct Answer - c The acceleration of a body undergoing uniform circular motion is called centripetal acceleration, and it is always directed the centre. |
|
| 113. |
The average acceleration vector for a particle having a uniform circular motion isA. A constant vector of magnitude `(v^(2))/(r )`B. A vector of magnitude `(v^(2))/(r)` directed normal to the plane of the given uniform circular motionC. Equal to the instantaneous acceleration vector at the start of the motionD. A null vector |
| Answer» Correct Answer - D | |
| 114. |
Two masses M and m are attached to a vertical axis by weightless threads of combined length l . They are set in rotational motion in a horizontal plane about this axis with constant angular velocity `omega` . If the tensions in the threads are the same during motion, the distance of M from the axis isA. `(Ml)/(M+m)`B. `(ml)/(M+m)`C. `(M+m)/(M)l`D. `(M+m)/(m)l` |
| Answer» Correct Answer - B | |
| 115. |
A boy on a cycle pedals around a circle of 20 metres radius at a speed of `20 "metres"//sec`. The combined mass of the boy and the cycle is 90 kg . The angle that the cycle makes with the vertical so that it may not fall is `(g = 9.8 m//sec^(2))`A. `60.25^(@)`B. `63.90^(@)`C. `26.12^(@)`D. `30.00^(@)` |
| Answer» Correct Answer - B | |
| 116. |
In the previous problem, the velocity of the shell will beA. `30ms^(-1)`B. `30sqrt(2)ms^(-1)`C. `30sqrt(3)ms^(-1)`D. none of these |
|
Answer» Correct Answer - b `45=H=(u^(2) sin^(2)45^(@))/(2xx10)implies 45=(u^(2).1//2)/(2xx10)` `u^(2)=45xx20xx2=90xx20impliesu=30sqrt2` |
|
| 117. |
A particle is tied to 20 cm long string. It performs circular motion in vertical plane. What is the angular velocity of string when the tension in the string at the top is zeroA. `5 rad//sec`B. `2 rad//sec`C. `7.5 rad//sec`D. `7 rad//sec` |
| Answer» Correct Answer - D | |
| 118. |
The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio `5:3` then its velocity isA. `sqrt(98) m//s`B. `7m//s`C. `sqrt(490) m//s`D. `sqrt(4.9)` |
| Answer» Correct Answer - A | |
| 119. |
For a particle performing uniform circular motion, choose the incorrect statement form the following.A. Magnitude of particle velocity (speed) remains constant.B. Particle velocity remains directed perpendicular to radius vector.C. Direction of acceleration keeps changing as particle moves.D. Magnitude of acceleration does not remain constant. |
|
Answer» Correct Answer - d For a particle performing uniform circular motion, magnitude of the acceleration remains constant. |
|
| 120. |
Which of the following statements is incorrect?A. In one dimension motion, the velocity and the acceleration of an object are always along the same line.B. In two or three dimension, the angle between velocity and acceleration vectors may have any value of between `0^(@) and 180^(@)`C. The kinematic equations for uniform acceleration can applied in case of a uniform circular motion.D. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. |
|
Answer» Correct Answer - c The kinematic equations for uniform acceleration do not apply in case of uniform circular motion becuase in this the magnitude of acceleration is constant but its direction is changing. |
|
| 121. |
A particle moves in xy plane. The rate of changes of `theta` at time t=2 second ( where `theta` is the angle which its velocity vector maks with positive x-axis) isA. `2/17rad//s`B. `1/14 rad//s`C. `4/7rad//s`D. `6/5rad//s` |
|
Answer» Correct Answer - a `x=2timplies v_(x)=(dx)/(dt)=2` `Y=2t^(2)implies v_(y)=(dy)/(dt)=4t` `:. Ttan theta =(v_(y))/(v_(x))=(4t)/2=2t` Differentiating with respect to time we get, `(sec^(2) theta)(d theta)/2=2` or `(1+tan^(2) theta) (d theta)/2=2, or (1+4t^(2)) (d theta)/(dt)=2` or `(d theta)/(dt)=2/(1+4t^(2)), (d theta)/(dt) at t=2s` is `(d theta)/(dt)=2/(1+4(2)^(2))=2/17 rad//s` |
|
| 122. |
A particle (A) is dropped from a height and another particles (B) is thrown into horizontal direction with speed of 5m/s sec from the same height. The correct statement isA. Both partiles will reaches at ground simultaneouslyB. Both particles will reaches at ground with the same speedC. Particle (A) will reach at ground at first with respect to particle (B)D. Particle (B) will reach at ground at first with respect to particle (A) |
|
Answer» Correct Answer - a For both cases `t=sqrt((2h)/g)` =constant. Because vertical downward component of velocity will be zero for both the particles. |
|
| 123. |
Which out of these does not affect the maximum height of a projectile?A. Mass of projectileB. Angle of projectionC. Acceleration due to gravityD. Magnitude of initial velocity |
|
Answer» Correct Answer - a `H_(max)=(u^(2))/(2g)sin^(2) theta` `H_(max)` is the independent of mass. |
|
| 124. |
In a two dimensional motion,instantaneous speed `v_(0)` is a positive constant.Then which of the following are necessarily true?A. The acceleration of the particle is zero.B. The acceleration of the particle is boundedC. The acceleration of the particle is necessarily in the plane of motion.D. The particle must be undergoing a uniform circular motion. |
|
Answer» Correct Answer - c In two dimensional motion, if instantaneous speed is a positive constan, then the accleration of the particle is necessarily in the plane of motion. |
|
| 125. |
If the range of a gun which fires a shell with muzzle speed V is R , then the angle of elevation of the gun isA. `cos^(-1)((V^(2))/(Rg))`B. `cos^(-1)((gR)/(V^(2)))`C. `(1)/(2)((V^(2))/(Rg))`D. `(1)/(2)sin^(-1)((gR)/(V^(2)))` |
| Answer» Correct Answer - D | |
| 126. |
In the previous problem, the maximum horizontal distance will beA. `160sqrt(3)m`B. `140sqrt(3)m`C. `120sqrt(3)m`D. `100sqrt(3)m` |
|
Answer» Correct Answer - a Horizontal range, `R=(u^(2)sin 2 theta)/g` `R=((56)^(2)sin 60^(@))/(9.8)` `=(56xx56xxsqrt3)/(9.8xx2)=160sqrt3m` |
|
| 127. |
For angles of projection of a projectile at angle `(45^(@) - theta) and (45^(@)+ theta)`, the horizontal ranges described by the projectile are in the ratio of :A. `2:1`B. `1:2`C. `1:1`D. `2:3` |
|
Answer» Correct Answer - c For a projectile launched with velocity u at an angle `theta`, the horizontal range is given by `R=(u^(2)sin 2 theta)/g` (i) for `theta_(1)=45^(@)- theta` `R_(1)=(u^(2)sin (90^(@)-2 theta))/g=(u^(2)cos 2 theta)/g` (ii) For `theta_(2)=45^(@)+ theta` |
|
| 128. |
A projectile fired with initial velocity u at some angle `theta` has a range R . If the initial velocity be doubled at the same angle of projection, then the range will beA. `2R`B. `R//2`C. `R`D. `4R` |
| Answer» Correct Answer - D | |
| 129. |
Assertion: When range of a projectile is maximum , its angle of projection may be `45^(@) or 135^(@)`. Reason: Whether `theta is 45^(@) or 135^(@)`, value of range remains the same, only the sign changes.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 - A | |
| 130. |
Assertion: When range of a projectile is maximum , its angle of projection may be `45^(@) or 135^(@)`. Reason: Whether `theta is 45^(@) or 135^(@)`, value of range remains the same, only the sign changes.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. |
|
Answer» Correct Answer - a Range, `R=(u^(2)sin2 theta)/g` When `theta=45^(@), R_(max)=(u^(2))/g sin 90^(@)=(u^(2))/g` When `theta=135^(@), R_(max)=(u^(2))/g sin 270^(@)=(-u^(2))/g` Negative sign shows opposite direction. |
|
| 131. |
Assertion: Horizontal range is same for angle of projection `theta and (90^(@)- theta)`. Reason : Horizontal range is independent of angle of projection.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. |
|
Answer» Correct Answer - c Horizontal range depends upon angle of projection and it is same for complementary angles i.e. `theta and (90- theta)` |
|
| 132. |
A swimmer crosses a flowing stream of width `d` to and fro normal to the flow of the river at time `t_(1)`. The time taken to cover the same distance up and down the stream is `t_(2)`. If `t_(3)` is the time the swimmer would take to swim a distance `2d` in still water, then relation between `t_(1),t_(2)`&`t_(3)`.A. `t_(1)^(2)=t_(2)t_(3)`B. `t_(2)^(2)=t_(1)t_(3)`C. `t_(3)^(2)=t_(1)t_(2)`D. `t_(3)=t_(1)+t_(2)` |
|
Answer» Correct Answer - a Let v be the river velocity and u the velocity of swimmer in still water. Then `t_(1)=2(W/(sqrt(u^(2)-v^(2))))` `t_(2)=W/(u+v)+W/(u-v)=(2uW)/(u^(2)-v^(2)) and t_(3)=(2W)/u` Now we can see that `t_(1)^(2)=t_(2)t_(3)` |
|
| 133. |
An insect trapped in circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100s. The linear speed of the insect isA. `4.3 cms^(-1)`B. `5.3 cms^(-1)`C. `6.3 cms^(-1)`D. `7.3 cms^(-1)` |
|
Answer» Correct Answer - b Here, r=12 cm Frequency `V=7/100Hz` The angular speed of the insect is `omega=2piv=2pixx7/100=0.44 rads^(-1)` The linear speed the insect is `v=omegar=0.44xx12=5.3cms^(-1)` |
|
| 134. |
In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to. .A. `(ut)/(sqrt3)`B. `(sqrt(3) ut)/2`C. `sqrt3 ut`D. `2ut` |
|
Answer» Correct Answer - a Horizontal component of velocity, `u_(H)=ucos 60^(@)=u/2` `AC=u_(H)xxt=(ut)/2` And `AB=AC sec 30^(@)` n `=((ut)/2)(2/(sqrt(3)))=(ut)/(sqrt(3))` |
|
| 135. |
Assertion: Two particles of different mass, projected with same velocity at same angles. The maximum height attained by both the particle will be same. Reason: The maximum height of projetile is independent of particle mass.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. |
|
Answer» Correct Answer - a `H=(u^(2) sin^(2) theta)/(2g)` i.e. it is independent of mass of projetile. |
|
| 136. |
Assertion: Two particles of different mass, projected with same velocity at same angles. The maximum height attained by both the particle will be same. Reason: The maximum height of projetile is independent of particle mass.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 - A | |
| 137. |
Two particles of mass M and m are moving in a circle of radii R and r. if their time period are the same, what will be the ratio of their linear velocities?A. `MR:mr`B. `M:m`C. `R:r`D. `1:1` |
|
Answer» Correct Answer - c Linear velocity `v=romega` `v_(1)=omegar_(1), v_(2)=omegar_(2)` [ `omega` is the same in both cases because time periode is the same] `(v_(1))/(v_(2))=(r_(1))/(r_(2))=R/r` |
|
| 138. |
Four persons K,L,M,N are initially at the four corners of a square of side d. Each person now moves with a uniform speed v in such a way that K always moves directly towards L, L directly towards M, M directly towards N, and N directly towards K. The four persons will meet at a time............. .A. `(d)/(v) sec`B. `(sqrt(2d))/(v) sec`C. `(d)/(sqrt(2v))sec`D. `(d)/(2v) sec` |
| Answer» Correct Answer - A | |
| 139. |
The coordinate of a particle moving in a plane are given by ` x(t) = a cos (pt) and y(t) = b sin (pt)` where `a,b (lt a)` and `P` are positive constants of appropriate dimensions . ThenA. The path of the particle is an ellipseB. The velocity and acceleration of the particle are normal to each other at `t = pi//(2p)`C. The acceleration of the particle is always directed towards a focusD. The distance travelled by the particle in time interval `t = 0` to `t = pi//(2p)` is a |
| Answer» Correct Answer - A::B | |
| 140. |
At what angle with the horizontal should a ball be thrown so that the range `R` is related to the time of flight as `R = 5 T^2` ? `(Take g = 10 ms6-2)`.A. `60^(@)`B. `30^(@)`C. `45^(@)`D. `75^(@)` |
|
Answer» Correct Answer - a `R=5T^(2)` `implies (u^(2)sin2 theta)/g=(5(u^(2)sin2 theta)^(2))/(g^(2))` `implies 2u^(2) sin theta cos theta=(20(u^(2)sin^(2) theta))/g` `implies tan theta=1 implies theta=45^(@)` |
|
| 141. |
An aircraft executes a horizontal loop with a speed of 150 m/s with its, wings banked at an angle of `12^(@)`. The radius of the loop is `(g=10m//s^(2))`A. `10.6 km`B. 9.6 kmC. 7.4 kmD. 5.8 km |
| Answer» Correct Answer - A | |
| 142. |
The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are (0,2). The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are A. (1,4)B. (5,3)C. (3, 4)D. (4, 1) |
| Answer» Correct Answer - B | |
| 143. |
Assertion: A projectile that traverses a parabolic path show deviation from its idealised trajectory in the presence of air resistance. Reason: Air resistance affect the motion of the projectile.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. |
|
Answer» Correct Answer - a In the presence of dissipative force like air friction which opposes the motion, any object will loss some part of its initial energy and consequently momentum too. Thus a projectile that traverses a parabolic path would certainly show deviation from its idealized trajectory in the presence of air resistance. |
|
| 144. |
The path of a projectile in the absence of air drag is shown in the figure by dotted line. If the air resistance is not ignored then which one of the paths shown in the figure is appropriate for the projectile? A. `B`B. `A`C. `D`D. `C` |
|
Answer» Correct Answer - a If air resistance is taken into consideration then range and maximum height, both will decreases. |
|
| 145. |
The path of a projectile in the absence of air drag is shown in the figure by dotted line. If the air resistance is not ignored then which one of the paths shown in the figure is appropriate for the projectile? A. BB. AC. DD. C |
| Answer» Correct Answer - A | |
| 146. |
Assertion: The maximum horizontal range of projectile is proportional to square of velocity. Reason: The maximum horizontal range of projectile is equal to maximum height attained by projectile.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. |
|
Answer» Correct Answer - c `R=(u^(2) sin 2 theta)/g :. R_(max)=(u^(2))/g when theta=45^(@) :. R_(max) prop u^(2)` Height `H=(u^(2)sin^(2) theta)/(2g) implies H_(max)=(u^(2))/(2g) when theta=90^(@)` It is clear that `H_(max)=(R_(max))/2` |
|
| 147. |
The path of one projectile as seen by an observer on another projectile is a/an:A. straight lineB. parabolaC. ellipseD. circle |
|
Answer» Correct Answer - a We know that `x=(u cos theta)t and y =(u sin theta)t-1/2"gt"^(2)` Let `x_(2)-x_(1)=(u_(1)cos theta_(1)-u_(2)cos theta_(2))t=X` `y_(2)-y_(1)=(u_(1)cos theta_(1))t-1/2"gt"^(2)-(u_(2)cos theta_(2))t+1/2"gt"^(2)` `=(u_(1)sin theta_(1)-u_(2)sin theta_(2))t=Y` `Y/X=((u_(1)sin theta_(1)-u_(2)sin theta_(2))t)/((u_(1)cos theta_(1)-u_(2)cos theta_(2))t)` `=(u_(1)sin theta_(1)-u_(2)sin theta_(2))/(u_(1)cos theta_(1)-u_(2)cos theta_(2))` =constant m(say) `Y=mX` It is equation of a straight line passing through the origin. |
|
| 148. |
A body is projected at `30^(@)` with the horizontal. The air offers resistance in proportional to the veclocity of the body. Which of the following statements is correct?A. The trajectroy is a symmetrical parabolaB. the time of rise to the maximum height is equal to the time of return to the groundC. The velocity at the highest pint is directed along the horizontalD. the sum of the kinetic and potential energies remains constant |
|
Answer» Correct Answer - c The upward motion is with higher retardation while the downward motion is with lesser acceleration. Further, the time of rise is less than the time of return. A part of the kinetic energy is used against friction. |
|
| 149. |
A helicopter is flying horizontally at `8m//s` at an altitude 180m when a package fo emergency medical supplies is ejected horizontally backward with a speed fo `12m//s` relative to the helicopter. Ignoring air resistance what is horizontal distance between the package and the helicopter when the package hits the ground?A. `120m`B. `24m`C. `36m`D. `72m` |
|
Answer» Correct Answer - d Time taken by package to reach the ground: `T=sqrt((2xx180)/10)=6s` Horizontal distance travelled by helicopter in this time=`8xx6=48m`. Velocity of package w.r.t ground `=12-8=4 m//s`in backward direction. Horizontal distance travelled by package in time `T=4xx6=24m`. So horizontal distance between them `=48+24=72m` |
|
| 150. |
From the top of a tower of height 40m, a ball is projected upward with a speed of `20ms^(-1)` at an angle of elevation of `30^(@)`. Then the ratio of the total time taken by the ball to hit the ground to the time taken to ball come at same level as top of tower.A. `2:1`B. `3:1`C. `3:2`D. `4:1` |
|
Answer» Correct Answer - a If t is the total time taken, then `40=-20sin 30^(@)t+1/2xx10xxt^(2)` or `40=-10t+5t^(2)` or `5t^(2)-10t-40=0` or `t^(2)-2t-8=0` or `t^(2)-4t+2t-8=0` or `t(t-4)+2(t-4)=0` or `(t+2)(t-4)=0` t=4s [Negative time is not allowed] `T=(2v sin theta)/g=(2xx20sin 30^(@))/10s=2s` `:. t/T=4/2=2/1` |
|