Physics(Exams > Bitsat ) Questions and answers for exam Preparation

Exams > Bitsat

PHYSICS MCQs

Total Questions : 421 | Page 7 of 43 pages

Question 61.

In previous problem, find work done by lower block on upper block in the frame of lower block

- 1 joule
- 2 joule
2 joule
zero

Discuss Question

Answer: Option D. -> zero

The relative displacement of upper block with respect to lower block is zero. Hence work done on upper block in the frame of lower block will be zero.

Question 62.

If a man having bag in his hand moves up on a stair . Then

the work done by lifting force is zero
the work done by lifting force is non -zero with respect to ground.
the work done by lifting force is zero with respect to ground
the work done with respectt to ground is same as that of with respect to him.

Discuss Question

Answer: Option B. -> the work done by lifting force is non -zero with respect to ground.

The displacement of bad is in opposition of force of gravity , So work is done by lifting force against gravitational force .

Question 63.

A body of mass 1 kg moves from point A(2 m, 3 m, 4 m) to B ( 3 m, 2 m, 5 m) . During motion of body , a force

`vec(F) = (2n) hat(i) - (4N) hat(j )` acts on it. The work done by the force on the particle during displacement is

`2 hat(i) - 4 hat (j)`
2 joule
- 2 joule
none of these.

Discuss Question

Answer: Option C. -> - 2 joule

The displacement of body is

`vec(AB) = vec (r_B) - vec( r_A)`

= `(3 hat(i) + 2 hat (j) + 5hat(k)) - (2hat (i) + 3hat (j) + 4hat(k))`

= `hat(i) - hat(j) + hat (k)`

`:.` W = `vec(F) * vec(AB)`

= `(2 hat(i) - 4hat(j)) * ( hat(i) - hat(j) + hat(k))`

= 2 - 4 = - 2 joule.

Question 64.

Work done during raising a box on to a platform

depends upon how fast it is raised
does no depend upon how fast itis raised
does not depend upon mass of the box
both (A) and (B) are correct

Discuss Question

Answer: Option B. -> does no depend upon how fast itis raised

The work done by constant force does no depend upon path followed by body . It depends only upon initial and final position . Hence , (B) is correct .

Question 65.

In previous problem Q. (1) , the work done by normal reaction is

20 joule
196 joule
zero
none of these

Discuss Question

Answer: Option C. -> zero

`because ` w =` vec(F) * vec(s) = vec(N) * vec(s)`

= Ns cos `90^circ = 0`

Question 66.

A body of mass 10 kg is moving on a horizontal surface by applying a force of 10 N in forward direction . If body moves with constant velocity . Find the work done by applied force for a displacement of 2 m ?

20 joule
10 joule
30 joule
40 joule

Discuss Question

Answer: Option A. -> 20 joule

The work done by constant force is given by

` w = vec(F) * vec(s) = Fs cos theta = Fs cos 0`

= 10 x 2 = 20 joule.

Question 67.

A particle of mass m is moving in a horizontal circle of radius r under a centripetal force given by `( -k/r^2)`, where k is a constant, them

the total energy of the particle is `-((k)/(2r))`
the kinetic energy of the particle is `(k/r)`
the potential energy of the particle is `((k)/(2r))`
the kinetic energy of the particle is `( - k/r)`

Discuss Question

Answer: Option A. -> the total energy of the particle is `-((k)/(2r))`

Centripetal force = `(mv^2)/(r) = k/r^2`

`:.` `mv^2 = k/r`

`:.` Kinetic energy is `1/2 mv^2 = (k)/(2r)`

`:.` `F = - (k)/(2r), F = (du)/(dr)`

At infinity , potential energy is zero

`:.` `int_u^0 du = - int _r^oo Fdr`

`:.` `u = - k [ - 1/r]_r^oo = - k/r`

`:.` Total energy = `K.E. + u`

= `(k)/(2r) - k/r = - (k)/(2r)`

Question 68.

A particle moves on a curved path with constant speed v. The acceleration of the particle at `x` = 0 is `2av^2` . he path of particle is

dtraight line
parabolic
elliptcal
none of these

Discuss Question

Answer: Option B. -> parabolic

Question 69.

Kinetic energy of a particle moving along a circle of radius R depends on the distance covered as `T = ks^2` where k is a constant . Find the force acting on the particle as a function of s

`(2k)/(s) sqrt(1 + (s/R)^2)`
`2ks sqrt(1 + (R/s)^2)`
`2ks sqrt(1 + (s/R)^2)`
`(2s)/(k) sqrt(1 + (R/s)^2)`

Discuss Question

Answer: Option C. -> `2ks sqrt(1 + (s/R)^2)`

Question 70.

A particle of mass m is attached o one end of a string of length `l` while the other end is fixed to a point `h` above the horizontal table , the particle is made to revolve in a circle on the table , so as to make `p` revolution per second. The maximum value of p if the particle is to be in contact with the table will be :