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

Exams > Bitsat

PHYSICS MCQs

Total Questions : 421 | Page 8 of 43 pages

A person wants to drive on the vertical surface of a large cylindrical wooden well commonly known as death well in a circus. The radius of well is R and the coefficient of friction between the tyres of motorcycle and the wall of the well is `mu` .The minimum speed , the motorcyclist must have in order to prevent slipping should be

`sqrt(((Rg)/(mu_s)))`
`sqrt(((mu_s)/(Rg)))`
`sqrt(((mu_s g)/(R)))`
`sqrt(((R)/(mu_s g)))`

Discuss Question

Answer: Option A. -> `sqrt(((Rg)/(mu_s)))`

` N= (mv^2)/(R)`

`mu_2 N > mg`

`(mu_2 mv^2)/(R) > mg`

or `v > sqrt(((Rg)/(mu_s)))`

`:.` `v_min = sqrt(((Rg)/(mu_s)))`

Question 72.

A car runs from east to west and another car B of the same mass runs from west to east at the same path along the equator . A presses the track with a force `N_1` and B presses the track with a force `N_2`. Then

`N_1 > N_2`
`N_1 < N_2`
`N_1 = N_2`
none of these

Discuss Question

Answer: Option A. -> `N_1 > N_2`

Question 73.

A car of 400 kg is moving on a circle path of radius 30 m with a speed of 40 km/h . When the driver applies he braes and the car continues to move along the circular path , what is the maximum deceleration possible if the tyres are limited to a total horizontal friction of 10.6 kN ?

10 m/`s^2`
6.36 m/`s^2`
4 m/`s^2`
None of these

Discuss Question

Answer: Option B. -> 6.36 m/`s^2`

The net acceleration to car is provided by force of friction .

`:.` `a = sqrt(a_n^2 + a_t^2)`

`:.` `F = ma = m sqrt((v^2/r)^2 + a_t^2)`

or ` 10.6 xx 10^3 = m sqrt( v^4/r^2 + a_t^2)`

After solving , `a_t `= 6.36 m / `s^2`

Question 74.

A road of width 20 m forms an arc of radius 15 m, its outer edge is 2 m higher than its inner edge. Calculate for what velocity the road is banked.

`sqrt(10)` m/s
`sqrt14.7` m/s
`sqrt(9.8)` m/s
none of these

Discuss Question

Answer: Option B. -> `sqrt14.7` m/s

Condition for velocity on banked road

`because tan theta = v^2/rg`

`:.` ` v = sqrt(rg tan theta)`

But tan` theta = theta = sin theta = 2/20` ( for small angle)

`:.` `v = sqrt(15 xx 9.8 xx 2/20)`

= `sqrt(14.7)` m/s

Question 75.

A cyclist is travelling on a circular section of highway of radius 2500 ft at the speed of 60 mile / h. The cyclist suddenly applies the brakes causing the bicycle to slow down at constant rate . Knowing that after eight second , the speed has been reduced to 45 mile /h . The acceleration of the bicycle immediately after the brakes have been applied is :

2 ft/`s^2`
4.14 ft/`s^2`
3.10 ft/`s^2`
2.75 ft/`s^2`

Discuss Question

Answer: Option B. -> 4.14 ft/`s^2`

Tangential acceleration ` a_1` = `(v_f - v_i)/(t)`

=` ((45 - 60)/(8)) (22/15)` ft/s

= `- 11/4` ft /`s^2`

The radial acceleration is `a_n` = ` v^2/r`

= `( (60 xx 22/15)^2)/(2500)` = 3.1 ft/`s^2`

`:.` `a = sqrt(a_n^2 + a_t^2)` = 4.14 ft / `s^2`

Question 76.

Find the maximum speed with which an automobile can round a curve of radius 8 m without slipping if the road is unbanked and the co-efficient of friction between the road and the tyres is 0.8(g = 10 m/`s^2`)

8 m/s
10 m/s
20 m/s
none of these

Discuss Question

Answer: Option A. -> 8 m/s

Here centripetal force is provided by force of friction .

`because f = mu_s N`

`:.` ` f = (mv^2)/(r)`

or `mu_s N = (mv^2)/(r)`

or ` mu_s mg = (mv^2)/(r)`

`:.` ` v = sqrt(mu gr) = sqrt(0.8 xx 10 xx 8) ` = 8 m/s

Question 77.

A cyclist goes round a circular path of length 400 m in 20 second . calculate the angle through which he bends from vertical in order to maintain the balance.

`sin^-1 (0.64)`
`tan ^-1 (0.64)`
`cos ^1 (0.64)`
None of these .

Discuss Question

Answer: Option B. -> `tan ^-1 (0.64)`

` 400 = 2 pi r`

`because tan theta = v^2/rg`

= ` ((400/20)^2)/((400/2 pi)g)`

`tan theta = (400 xx 2 pi)/(400 g) = (6.28)/(9.8)` = 0.64

`because theta = tan ^-1 (0.64)`

Question 78.

Two buses A and B are moving around concentric circular paths of radii `r_A` and `R_B` If the two buses complete the circular paths in the same time . The ratio of their linear speeds is :

1
`r_A/r_B`
`r_B/r_A`
None of these.

Discuss Question

Answer: Option B. -> `r_A/r_B`

In this time period of both buses are same.

Here `T_1 = (2pi)/(omega_1)`

`T_2 = (2pi)/(omega_2)`

`because omega_1 = omega_2` `(because T_1 = T_2)`

`:.` `v_A/v_B = (r_A omega)/(r_B omega) = r_A/r_B`

Question 79.

A solid body rotates about a stationary axis so that its angular velocity depends on the rotational angle `phi` as `omega = omega_0 - k phi` where `omega_0` and k are positive constants . At the moment t = 0, `phi ` = 0. Find the time dependence of rotation angle :