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

Exams > Bitsat

PHYSICS MCQs

Total Questions : 421 | Page 10 of 43 pages

A uniform fine chain of length `l` is suspended with lower end just touching a horizontal table . Find the pressure on the table, when a length `x` has reached the table .

`mgx`
`2mgx`
`3mgx`
`(mgx)/(2)`

Discuss Question

Answer: Option C. -> `3mgx`

(C) Let m he mass per unit length of chain . When the velocity is `v` then a length `vdt` is brought to rest on the table in a short interval of time dt i.e. , a mass `mvdt` i.e. a brought to rest . Hence , amount of momentum changed in time

`dt = mvdt . v = mv^2dt`.

The force on the table due to the rate of change of falls freely , the velocity , when a length `x` has reached the table is given by `v^2 = 2gx`

Since , the weight of length `x` of the chain is `mgx`.

`:.` Total pressure = `mgx + mv^2`

= `mgx + 2mgx = 3mgx`

Question 92.

A block of mass m is moving on a wedge with the acceleration `a_0`. he wedge is moving with the acceleration `a_1`. The observer is situated on wedge . The magnitude of pseudo force on the block is :

`ma_0`
`ma_1`
`msqrt(a_0^2 + a_1^2)`
`m((a_1 + a_0)/(2))`

Discuss Question

Answer: Option B. -> `ma_1`

The magnitude and direction of pseudo force depends upon acceleration of reference frame in which observer is situated . It does not depend upon direction and magnitude of acceleration of the block .

So, (B) is correct.

Question 93.

A body of mass 10 kg is to be raised by a massless string from rest to rest , through a height 9.8 m.The greatest tension which he string can safely bear is 20 kg wt. The least time of ascent is

1 sec
3 sec
4 sec
none of these

Discuss Question

Answer: Option A. -> 1 sec

`a_max = (T - mg)/(m) =( (20 - 10)g)/(10) = g`

During half time , body is accelerated from rest and during next half time decelerates with same magnitude to come to rest.

`:.` `h/2 = 1/2 a_max t_1^2`

`:.` `t_1 = sqrt(h/a_max)` = 1 sec.

Question 94.

A man is pulling a rope attached to a block on a smooth horizontal table . The tension in the rope will be the same at all points .

If and only if the rope is not accelerated
if and only if the rope is massless
if either the rope is not accelerated or is massless
always

Discuss Question

Answer: Option C. -> if either the rope is not accelerated or is massless

Question 95.

A particle of mass m moves on the `x`-axis under the influence of a force of attraction towards the origin O given by

`F = - k/x^2``vec(i)`. If the particle starts from rest at `x` = a . The speed it will attain to reach at distance `x` from the origin O will be .

`sqrt(2k)/m [(x - a)/(ax)]^(1/2)`
`sqrt(2k)/m [(a + k)/(ax)]^(1/2)`
`sqrt(k)/m [(ax)/(x - a)]^(1/2)`
`sqrt(m)/2k [(a - x)/(ax)]^(1/2)`

Discuss Question

Answer: Option A. -> `sqrt(2k)/m [(x - a)/(ax)]^(1/2)`

`because` `F = k/x^2`

`:.` Acceleration `int = - (k)/(mx^2)`

when `x` secreases , v increases

`:.` `int = - v (dv)/(dx)`

`:.` ` - v (dv)/(dx) = (k)/(mx^2)`

or `int_0^v vdv = k/m int_a^x 1/x^2 dx`

`:.` v = `sqrt(2k)/m [(x - a)/(ax)]^(1/2)`

Question 96.

Choose the correct option or options

tension force always pulls a body
tension can never push a body or rope
tension across massless or frictionless pulley remains constant
ropes become slack when ension force becomes zero
all the above

Discuss Question

Answer: Option D. -> ropes become slack when ension force becomes zero

Question 97.

The action and reaction forces referred to Newton's third law of motion.

must acr upon the same body ,
must act upon different bodies
need not to be equal in magnitude but must have the same line of action
must be equal in magnitude but need not have the same line of action.

Discuss Question

Answer: Option B. -> must act upon different bodies

Question 98.

A o.1 kg body moves at a constant speed of 10 m/s. It is pushed by applying a constant force for 2 sec. Due to this force , it starts moving exactly in the opposite direction with a speed of 4 m/s ./ Then

the deceleration of the body is 7 m/`s^2`
the magnitude of change in momentum is 1.4 kg m/sec
impulse of the force is 1.4 Ns
the force which acts on the ball is 0.7 N
all the above

Discuss Question

Answer: Option E. -> all the above

`a = (v - u)/(t) = (-4 - 10)/(2)` = - 7 m/`s^2`

The magnitude of change in momentum

= `[m(v - u)]`

= 1.4 kg m/s = Impulse

Now, F = ma = 0.7 N

Question 99.

Water jet issues water from a nozzle of 2 `cm^2` cross-section with velocity 30 cm/s and strikes a plane surface placed at right angles to the jet . The force exerted on the plane is .

200 dyne
400 dyne
1800 dyne
none of these

Discuss Question

Answer: Option C. -> 1800 dyne

`because` `F = (Delta P)/(Delta t) = v (Delta m)/(Delta t)`

= `pAv (Delta x)/(Delta t)`

= `pAv ((v Delta t)/(Delta t))` = 1 x 2 x `(30)^2`

= 1800 dyne.

Question 100.

A particle of mass m moves on the `x`-axis as follows. I starts from rest at t = 0 from the point `x`= 0 and comes to rest at t = 1 at the point `x` = 1. No other information is available about its motion at intermediate time (0 < t < 1). If `alpha` denotes the instantaneous acceleration of the particle , then.

`alpha` cannot remain posiive for all t in the interval `0 le t le 1`
`[alpha]` cannot exceed 2 a any point in its path
`[alpha]` be `ge` 4 at point or some points in its path
both (A) and (C) are correct .

Discuss Question

Answer: Option D. -> both (A) and (C) are correct .

Since at t = 0 and t = 1 sec, particle is in rest . This is possible only when the particle accelerates for some time and then decelerates to come in rest.