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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. |
Assertion : In circular motion, the centripetal and centrifugal force acting in opposite direction balance each other. Reason : Centripetal and centrifugal forces don’t act at the same time.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 | |
| 52. |
Asseration : In circular motion work done by all the forces acting on the body is zero. Reason : Centripetal force and veloity are mutually perpendicular.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 | |
| 53. |
The length of a seconds hand in watch is `1 cm.` The change in velocity of its tip in `15 s` isA. zeroB. `(pi)/(30sqrt(2)) cm//sec`C. `(pi)/(30) cm//sec`D. `(pi sqrt(2))/30 cm//sec` |
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Answer» Correct Answer - d `Deltav=2v sin ((90^(@))/2)` `=2v sin 45^(@)=2vxx1/(sqrt(2))=sqrt(2)v` `=sqrt(2)xxr omega=sqrt(2)xx1xx(2pi)/60=(sqrt(2)pi)/(30) cm//s` |
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| 54. |
Assertion: Centripetal acceleration is always direction towards the centre of rotation of an object undergoing uniform circular motion Reason: Centripetal acceleration is a constant vectorA. 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 - c Centripetal accleration `=(v^(2))/R` For uniform circular motion, v and R are constant, the magnitude of the centripetal acceleration is constants. But the direction change-pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector. |
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| 55. |
The ratio of angular speeds of minute hand and hour hand of a watch isA. `1:12`B. `6:1`C. `12:1`D. `1:6` |
| Answer» Correct Answer - C | |
| 56. |
The angle turned by a body undergoing circular motion depends on time as `theta = theta_(0)+theta_(1)t+theta_(2)t^(2)`. Then the angular acceleration of the body isA. `theta_(1)`B. `theta_(2)`C. `2theta_(1)`D. `2theta_(2)` |
| Answer» Correct Answer - D | |
| 57. |
Two balls `A and B` are thrown with speeds `u and u//2`, respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the angle of projection of ball `B is 15^@` with the horizontal, then the angle of projection of `A` is.A. `sin^(-1)(1/8)`B. `1/2sin^(-1)(1/8)`C. `1/3sin^(-1)(1/8)`D. `1/4sin^(-1)(1/8)` |
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Answer» Correct Answer - b `(u^(2)sin2 theta)/g=((u//2)^(2) sin 30^(@))/g=(u^(2))/(8g)` `:. sin 2 theta=1/8` or `theta=1/2 sin^(-1)(1/8)` |
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| 58. |
Two point particles with masses `m_(1) and m_(2)` are thrown at angles `theta_(1) and theta_(2)` with horizontal with speeds `v_(1) and v_(2)` respectively. R, H and T are range, height and total time of flight respectively. Let `v_(1) sin theta_(1) =v_(2) sin theta_(2)`. Then for both particlesA. T,H and R are differentB. H and R will be same but T will be differentC. T and R are same but H will be differentD. T and H are same but R is different. |
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Answer» Correct Answer - d Relations for, `T=(2u sin theta)/g, R=(u^(2)sin 2 theta)/g and H=(u^(2)sin^(2) theta)/(2g)` Here, `u_(1)sin theta_(1)=u_(2) sin theta_(2)` T and H are same but will not be same. |
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| 59. |
A motor car travelling at `30 m//s` on a circular road of radius `500m`. It is increasing its speed at the rate of `2 ms^(-2)`. What its accleration at that instant ?A. `1.8ms^(-2)`B. `2ms^(-2)`C. `3.8ms^(-2)`D. `2.7ms^(-2)` |
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Answer» Correct Answer - d Radial acceleration `a_(r)=(v^(2))/r=((30)^(2))/500=1.8ms^(-2)` Tangential acceleration at `=2m//sec^(2)` Resultant acceleration `a^(2)=a_(r)^(2)+a_(t)^(2)+2a_(r)a_(t)cos theta` Here, `theta=90^(@)` `a^(2)=(1.8)^(2)+(2)^(2)+0=3.24+4=7.24` `a=sqrt(7.24)=2.7 ms^(-2)` |
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| 60. |
A ball is rolled off the edge of a horizontal table at a speed of `4 m//"second"`. It hits the ground after 0.4 second . Which statement given below is trueA. It hits the ground at a horizontal distance 1.6 m from the edge of the tableB. The speed with which it hits the ground is 4.0 m/secondC. Height of the table is `0.8 m `D. It hits the ground at an angle of `60^(@)` to the horizontal |
| Answer» Correct Answer - A::C | |
| 61. |
Assertion: The trajectory of an object moving under the same accleration due to gravity can be straight line or a parabola depending on the initial conditions. Reason: The shape of the trajectory of the motion of an object is determined by the acceleration alone.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 - c The shape of the trajectory of the motion of an object is not determine by the acceleration alonge but also depends on the initial conditions of motion (initial position and initial velocity). |
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| 62. |
A ball is thrown from the top of tower with an initial velocity of `10ms^(-1)` at an angle of `30^(@)` with the horizontal. If it hits the ground of a distance of 17.3m from the back of the tower, the height of the tower is `(Take g=10ms^(-2))`A. `5m`B. `20m`C. `15m`D. `10m` |
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Answer» Correct Answer - d Here, `theta=30^(@), u=10ms^(-1)`, `R=17.3m, g=10ms^(-2)` For horizontal motion, `R=u cos thetat` `t=R/(u cos theta)=17.3/(10 cos 30^(@))=(17.3xx2)/(10xxsqrt(3))=(17.3xx2)/(10xx1.73)=2s` For vertical motions, `h=u sin thetat=1/2"gt"^(2)` `=10sin 30^(@)xx2-1/2xx10xx2^(2)` `=10-20=-10m`. Height of tower =10m. |
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| 63. |
A particle is projected with initial velocity of `hati+2hatj`. The eqaution of trajectory is `(take g=10ms^(-2))`A. `y=2x-15x^(2)`B. `y=2x-25x^(2)`C. `y=x-5x^(2)`D. `y=2x-5x^(2)` |
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Answer» Correct Answer - b Given, Horizontal component of initial velocity `u_(x)=2=usin theta` `tan theta=(u sin theta)/(u cos theta)=2/1=2` The equation of trajectory of projectile motion is `y=x tan theta-(gx^(2))/(2u^(2)cos^(2) theta)=xtan theta-(gx^(2))/(2(u_(x))^(2))` `y=x xx2 -(10xx x^(2))/(2(1)^(2))=2x-5x^(2)` |
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| 64. |
In horizontal level ground to projectile if at any instant velocity becomes perpendicular to initial velocity then what can you say about projection angle with horizontal.A. `theta=45^(@)`B. `thetage45^(@)`C. `thetale45^(@)`D. for any value of `theta` it is possible |
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Answer» Correct Answer - b Velocity at any time `vecv=vecu+vec(g)t` `implies vecv=u cos theta hati+(u sin theta-gt)hatj` Let the any time this velocity becomes perpendicular to initial velocity. The `vecv.vecu=0` Solve to get `t=u/(g sin theta)` Now t should be less then/equal to time of flight. So, `tgeT`. `u/(g sin theta)le(2u sin theta)/g implies sin^(2) thetage1/2` `implies sin thetage1/(sqrt(2)) implies thetage45^(@)` |
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| 65. |
A particle of mass m is projected with a velocity v making an angle of `45^@` with the horizontal. The magnitude of the angular momentum of the projectile abut the point of projection when the particle is at its maximum height h is.A. ZeroB. `mv^(3)//(4sqrt(2)g)`C. `mv^(3)//(sqrt(2)g)`D. `mv^(2)//2g` |
| Answer» Correct Answer - B | |
| 66. |
A particle projected with velocity u at angle `theta` with horizontal at t=0. What is the magnitude of change in the velocity of the particle when it is at maximum height?A. `(u cos theta)/2`B. `u cos theta`C. `u sin theta`D. none of these |
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Answer» Correct Answer - c `vecu_(i)=u cos theta hati+u sin theta hatj` `vecu_(f)=u cos theta hati` `Deltavecu=vecu_(f)-vecu_(i)=-u sin thetahatj` `|Deltavecu|=u sin theta` |
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| 67. |
A stone projected with a velocity u at an angle q with the horizontal reaches maximum heights `H_(1)`. When it is projected with velocity u at an angle `(pi/2-theta)` with the horizontal, it reaches maximum height `H_(2)`. The relations between the horizontal range R of the projectile, `H_(1) and H_(2)`, isA. `R=4sqrt(H_(1)H_(2))`B. `R=4(H_(1)-H_(2))`C. `R=4(H_(1)+H_(2))`D. `R=(H_(1)^(2))/(H_(2)^(2))` |
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Answer» Correct Answer - a `H_(1)=(u^(2)sin^(2) theta)/(2g)` `H_(2)=(u^(2)sin^(2)(90^(@)- theta))/(2g)=(u^(2)cos^(2) theta)/(2g)` `H_(1)H_(2)=(u^(2)sin^(2) theta)/(2g)/(u^(2)cos^(2) theta)/(2g)` `=((u^(2)sin 2 theta)^(2))/(16g^(2))=(R^(2))/16` `:. R=4sqrt(H_(1)H_(2))` |
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| 68. |
A body is projected up a smooth inclined plane with velocity V from the point A as shown in the figure. The angle of inclination is `45^(@)` and the top is connected to a well of diameter 40m. If the body just manages to across the well, what is the value of V? Length of inclined plane is `20sqrt(2)m`. A. `40ms^(-1)`B. `40 sqrt(2) ms^(-1)`C. `20ms^(-1)`D. `20 sqrt(2) ms^(-1)` |
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Answer» Correct Answer - d Angle of projection form B is `45^(@)`. As the body is able to cross the well of diameter 40m, hence or `R=(v^(2))/g or v=sqrt(gR)` `v=sqrt(10xx40)=20ms^(-1)` On the inclined plane, the retardation is: `gsin alpha=g sin 45^(@)=10/(sqrt(2))ms^(-2)` Using `v^(2)-u^(2)=2ax` `(20)^(2)-u^(2)=2xx(-10/(sqrt(2)))xx20sqrt(2)` `u=20sqrt(2)ms^(-1), i.e, V=20sqrt(2)ms^(-1)` |
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| 69. |
A smooth square platform ABCD is moving towards right with a uniform speed v. At what angle `theta` must a particle be projected from A with speed u so that it strikes the point B A. `sin^(-1)(u/v)`B. `cos^(-1)(v/u)`C. `cos^(-1)(u/v)`D. `sin^(-1)(v/u)` |
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Answer» Correct Answer - b Particle will strike the point B if velocity of particle with respect to platform is along AB or component of its relative along AD is zero. So, `u cos theta=v` or `theta=cos^(-1)(v/u)` |
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| 70. |
The maximum velocity (in `ms^(-1)`) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding isA. 60B. 30C. 15D. 25 |
| Answer» Correct Answer - B | |
| 71. |
A `1kg` stone at the end of `1m` long string is whirled in a vertical circle at a constant speed of `4m//s`. The tension in the string is `6N`, when the stone is at `(g=10m//s^(2))`A. Top of the circleB. Bottom of the circleC. Half way downD. None of the above |
| Answer» Correct Answer - A | |
| 72. |
A tube of length L is filled completely with an incomeressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity `omega.` The force exerted by the liquid at the other end isA. `(Mlomega^(2))/(2)`B. `M L omega^(2)`C. `(M L omega^(2))/(4)`D. `(M L^(2) omega^(2))/(2)` |
| Answer» Correct Answer - A | |
| 73. |
A particle is projected with a velocity `vecv=8hati+6hatj m//s`. The time after which it will starts moving perpendicular to its initial direction of motions isA. `0.5s`B. `1.25s`C. `1s`D. `5//3s` |
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Answer» Correct Answer - d Given `vecv_(1)=8hati+6hatjm//s^(2)` Acceleration due to gravity, `veca=-ghati=-10hatjm//s^(2)` Velocity after time t, `vecv_(2)=vecv_(1)+vec(a)t` `vecv_(2)=(8hati+6hatj)-10thatj=8hatj+(6-10t)hatj` Now `vecv_(1) and vecv_(2)` will be perpendicular if `vecv_(1).vecv_(2)=0` `(8hati+6hatj).[8hati+(6-10t)hatj]=0` `64+6(6-10t)=0` `64+36-60t=0` `60t=100t=5/3 sec` |
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| 74. |
A cricket ball is hit `30^(@)` with the horizontal with kinetic energy K. The kinetic energy at the highest point isA. ZeroB. `K//4`C. `K//2`D. `3K//4` |
| Answer» Correct Answer - D | |
| 75. |
If a ston s to at a point which is at a distance d away and at a height h avove the point from where the stone starts, then what is the value of initial speed u if the stone is lauched at angle `theta`? A. `g/(cos theta) sqrt(d/(2(dtan theta-h))`B. `d/(cos theta) sqrt(d/(2(dtan theta-h))`C. `sqrt((gd^(2))/((h cos^(2) theta)))`D. `sqrt((gd^(2))/((d-h)))` |
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Answer» Correct Answer - b `h=(u sin theta)t-1/2"gt"^(2)` `d=(u cos theta)t` `t=d/(u cos theta)` `h=u sin theta.d/(u cos theta)-1/2g.(d^(2))/(u^(2)cos^(2)theta)` `u=d/(cos theta)sqrt(g/(2(dtan theta-h)))` |
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| 76. |
The maximum range of a projectile is 500m. If the particle is thrown up a plane is inclined at an angle of `30^(@)` with the same speed, the distance covered by it along the inclined plane will be:A. `250m`B. `500m`C. `750m`D. `100m` |
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Answer» Correct Answer - b For the maximum range, `theta=45^(@)` `R=(u^(2)sin 2 theta)/g=(u^(2))/g sin 90^(@)=(u^(2))/g` or `500=(u^(2))/g` The distance covered along the inclined plane can be obtain using the equation `v^(2)-u^(2)=2as` or `0-u^(2)=2(-g sin 30^(@))s` or `s=(u^(2))/g=500m` |
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| 77. |
The ratio of the range of the particle and its maximum range in the inclined plane is:A. `1:2`B. `1:sqrt(2)`C. `1:sqrt(3)`D. `1:1` |
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Answer» Correct Answer - d `R=v_(0)cos 30^(@) T-1/2g sin 30^(@) T^(2)` `=10 (sqrt(3))/2-2/(sqrt(3))-1/2xx10xx1/2(2/(sqrt(3)))^(2)=20/3` `R_(max)=(v_(0)^(2))/(g(1+sin theta_(0)))` `=(10^(2))/(10(1+sin theta_(0)))` `=(10^(2))/(10(1+sin 30^(@)))=20/3` So, `R/(R_(max))=1` |
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| 78. |
A cylclist is riding with a speed of `27 km h^-1`. As he approaches a circular turn on the road of radius `80 m`, he applies brakes and reduces his speed at the constant rate of `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?A. `0.68 ms^(-2)`B. `0.86 ms^(-2)`C. `0.56 ms^(-2)`D. `0.76 ms^(-2)` |
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Answer» Correct Answer - b Here, `v=27kmh^(-1)=27xx5/18ms^(-1)` `v=15/2=7.5ms^(-1)` `r=80m` Centripetal acceleration, `a_(c)=(v^(2))/r` `a_(c)=((7.5ms^(-1))^(2))/(80m)=0.7 ms^(-2)` Tangential acceleration, `at=0.5 ms^(-2)` Magnitude of the net acceleration is `a=sqrt((a_(c))^(2)+(a_(t))^(2))=sqrt((0.7)^(2)+(0.5)^(2))=0.86 ms^(-2)` |
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| 79. |
When a body is thrown with a velocity u making an angle `theta` with the horizontal plane, the maximum distance covered by it in horizontal direction isA. `(u^(2) sin theta)/(g)`B. `(u^(2) sin 2theta)/(2g)`C. `(u^(2)sin 2theta)/(g)`D. `(u^(2) cos 2 theta)/(g)` |
| Answer» Correct Answer - C | |
| 80. |
A football player throws a ball with a velocity of 50 metre/sec at an angle 30 degrees from the horizontal. The ball remains in the air for `(g = 10m//s^(2))`A. `2.5 sec`B. `1.25 sec`C. `5 sec`D. `0.625 sec` |
| Answer» Correct Answer - C | |
| 81. |
A person standing on a moving truck, throws a stone vertically up relative to himself. To a person, standing on the ground, the stone appears to : (immediately after being thrown).A. Rise vertically up and come downB. Rise towards the rear of the truckC. Move along a parabolic pathD. Rise straight and forward but inclined to the direction of motion of truck |
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Answer» Correct Answer - c With respect to the man on ground, the ston has horizontal velocity (equal to that of the truck) as well as vertical velocity. So, it would appear to move along a parabolic path. |
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| 82. |
Assertion : When a vehicle takes a turn on the road, it travels along a nearly circular path. Reason : In circular motion, velocity of vehicle remains same.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 - C | |
| 83. |
Assertion: For motion in two or three diemensions, velocity and acceleration vecotrs must have any angle between `0^(@) and 90^(@)` between them. Reason: For such motion velocity and acceleration of an object is always in the opposite 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 - d In one diemension, the velocity and the acceleration of an object are always along the same straight line either in the same direction or in the opposite direction. However, for motion in two or three dimension velocity and acceleration vectors may have any angle between `0^(@) and 180^(@)` between them. |
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| 84. |
A particle moves along a circle of radius R =1m so that its radius vector `vecr` relative to a point on its circumference rotates with the constant angular velocity `omega=2rad//s`. The linear speed of the particle isA. `4m//s`B. `2m//s`C. `1m//s`D. `0.5m//s` |
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Answer» Correct Answer - a Angular velocity about centre =2 (angular velocity about any point on its circumference) or `omega=2(2)rad//s` or `omega=4 rad//s` Now `v=romega=(1)(4)=4m//s` |
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| 85. |
A particle moves in a circle of radius 25 cm at two revolutions per sec. The acceleration of the particle in `m//s^(2)` is:A. `pi^(2)`B. `8pi^(2)`C. `4pi^(2)`D. `2pi^(2)` |
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Answer» Correct Answer - c `r=25xx10^(-2)m, f= 2//sec` `omega=2pif=4pi rad//sec` Acceleration `=ometa^(2)r=(4pi)^(2)xx25xx10^(-2)` `=16xx25xx10^(-2)pi^(2)m//s^(2)` `=4pi^(2)m//s^(2)` |
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| 86. |
An object is projected at an angle of `45^(@)` with the horizontal. The horizontal range and the maximum height reached will be in the ratio.A. `1:2`B. `2:1`C. `1:4`D. `4:1` |
| Answer» Correct Answer - D | |
| 87. |
A tachometer is a device to measureA. Gravitational pullB. Speed of rotationC. Surface tensionD. Tension in a spring |
| Answer» Correct Answer - B | |
| 88. |
If two bodies are projected at `30^(@)` and `60^(@)` respectively, with the same velocity, thenA. Their ranges are sameB. Their heights are sameC. Their times of flight are sameD. All of these |
| Answer» Correct Answer - A | |
| 89. |
In a projectile motion, velocity at maximum height isA. `(u cos theta)/(2)`B. `u cos theta`C. `(u sin theta)/(2)`D. None of these |
| Answer» Correct Answer - B | |
| 90. |
Which of the following quantities remain constant during projectile motion?A. Initial velocity inclined to the horizontalB. Zero velocity at the highest pointC. Constant acceleration perpendicular to the velocityD. None of the above |
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Answer» Correct Answer - d In a projectile motion, it is not necessary that initial must be inclined to the horizontal. Further, only vertical component of velocity is zero at the highest point. Also, constant acceleration is perpendicular to velocity only at the highest point. |
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| 91. |
A ball is moving to and fro about the lowest point A of a smooth hemispherical bowl. If it is able to rise up to a height of 20 cm on either side of A , its speed at A must be (Take `= 10m//s`, mass of the body 5 g )A. `0.2 m//s`B. `2 m//s`C. `4 m//s`D. `4.5 ms` |
| Answer» Correct Answer - B | |
| 92. |
A body of mass m is moving in a circle of radius r with a constant speed v, The force on the body is `(mv^(2))/(r )` and is directed towards the centre what is the work done by the from in moving the body over half the circumference of the circle?A. `(mv^(2))/r xxpir`B. zeroC. `(mv^(2))/(r^(2))`D. `(pir^(2))/(mv^(2))` |
| Answer» Correct Answer - B | |
| 93. |
Two particles P and Q are located at distance `r_(p) and r_(Q)` respectively form the centre of rotating disc such that `r_(P)gtr_(Q)`. The disc is rotating with constant angular acceleration. We can sayA. both P and Q have the same accelerationB. both P and Q do not have the same accelerationC. P has greater acceleration than QD. Q has greater acceleration than P |
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Answer» Correct Answer - c Resultant acceleration, `a=sqrt(a_(r)^(2)+a_(t)^(2))=sqrt((omega^(2)r)^(2)+(r alpha)^(2))` `implies a=rsqrt((omega^(2))^(2)+(alpha)^(2))` As `omegan and alpha` will be same for both the particles P and Q, hence `a_(p)gta_(Q)( :. R_(P) gt r_(Q))` |
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| 94. |
A particle moves in a circle of radius 25 cm at two revolutions per sec. The acceleration of the particle in `m//s^(2)` is:A. `pi^(2)`B. `8 pi^(2)`C. `4 pi^(2)`D. `2pi^(2)` |
| Answer» Correct Answer - C | |
| 95. |
The acceleration of a train travelling with speed of `400m//s` as it goes round a curve of radius 160 m , isA. `1 km//s^(2)`B. `100 m//s^(2)`C. `10 m//s^(2)`D. `1m//s^(2)` |
| Answer» Correct Answer - A | |
| 96. |
Assertion : Two similar trains are moving along the equatorial line with the same speed but in opposite direction. They will exert equal pressure on the rails. Reason : In uniform circular motion the magnitude of acceleration remains constant but the direction continuously 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 assertion is false but reason is true |
| Answer» Correct Answer - D | |
| 97. |
When a body moves with a constant speed along a circleA. No work is done on itB. No acceleration is produced in the bodyC. No force acts on the bodyD. Its velocity remains constant |
| Answer» Correct Answer - A | |
| 98. |
A body of mass m moves in a circular path with uniform angular velocity. The motion of the body has constantA. AccelerationB. VelocityC. MomentumD. Kinetic energy |
| Answer» Correct Answer - D | |
| 99. |
In uniform circular motionA. Both velocity and acceleration are constantB. Acceleration and speed are constant but velocity changesC. Both acceleration and velocity changesD. Both acceleration and speed are constant |
| Answer» Correct Answer - C | |
| 100. |
For a particle in a non-uniform accelerated circular motionA. Velocity is radial and acceleration is transverse onlyB. Velocity is transverse and acceleration is radial onlyC. Velocity is radial and acceleration has both radial and transverse componentsD. Velocity is transverse and acceleration has both radial and transverse components |
| Answer» Correct Answer - D | |