Question

Wired Faculty App

A car of mass 1000 kg moves on a circular track of radius 20 m. If the coefficient of friction of 0.64. The maximum velocity with which the car can be moved is

11.2m/s

112m/s

$\frac{0.64 \times 20}{1000 \times 100} m / s$

$\frac{1000}{64 \times 20} m / s$

Solution

A.

11.2m/s

$V_{\max} = \sqrt{μrg} = \sqrt{0.64 \times 20 \times 9.8} where, μ - coefficient of friction V_{\max} = 11.2 m / s$

Question

Wired Faculty App

The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in $\sqrt{22}$ sec is (g=9.8 m/s)

15⁰

30⁰

60⁰

45⁰

Solution

D.

45⁰

.Given:-Speed of particle V= $\frac{S}{t}$ = $\frac{34.3}{\sqrt{22}}$

radius r= $\frac{S}{2 π}$ = $\frac{34.3}{2 π}$

tan $θ$ = $\frac{V^{2}}{rg} = \frac{\frac{(34.3)^{2}}{({\sqrt{22)}}^{2}}}{\frac{34.3}{2 π} \times 9.8}$ = $\frac{34.3 \times 2 \times \frac{22}{7}}{22 \times 9.8}$ = $\frac{68.6}{68.6} = 1$

tan $θ$ = 1 .....(tan45⁰=1)

Question

Wired Faculty App

The radius of earth is 6400 km and g = 10 m/s². In order that a body of 5 kg weight is zero at the equator, the angular speed is

$\frac{1}{800} rad / \sec$

$\frac{1}{1600} rad / \sec$

$\frac{1}{400} rad / \sec$

$\frac{1}{80} rad / \sec$

Solution

A.

$\frac{1}{800} rad / \sec$

Given:- radius of earth=6400km=6400 $\times$ 10³, g=10m/s²

g^'=0, m=0(weight is Zero at equator)

g=R $ω$ ² $\Rightarrow$ $ω$ ²= $\frac{g}{R}$

Angular speed $ω$ = $\sqrt{\frac{g}{R}}$ = $\sqrt{\frac{10}{6400 \times 10^{3}}}$ = $\frac{1}{800} rad / \sec$

Question

Wired Faculty App

The escape velocity for the earth is 11.2 km/sec. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

56.0 km/sec

280 km/sec

112 km/sec

56 km/sec

Solution

A.

56.0 km/sec

Escape velocity of earth V_e=11.2km/sec

v_escape= $\sqrt{\frac{2 GM}{R}} \propto \sqrt{\frac{M}{R}}$

Hence, $\frac{V_{P}}{V_{e}} = \sqrt{\frac{\frac{M_{p}}{R_{P}}}{\frac{M_{e}}{R_{e}}}} = \sqrt{\frac{M_{P}}{R_{p}} \times \frac{R_{e}}{M_{e}}}$

Or $\frac{V_{p}}{V_{e}} = \sqrt{100 \times \frac{1}{4}} = \sqrt{25} = 5$

V_p=5 $\times$ V_e=5 $\times$ 11.2 =56km/sec

Question

Wired Faculty App

A wheel rotates with constant acceleration of 2.0 radian/sec² .If the wheel starts from rest the number of revolution it makes in the first ten second, will be approximately

32

24

16

8

Solution

C.

16

Angular acceleration= 2 rad/s²

Time interval t= 10sec

Angular displacement θ = ?

$N ow by θ = ωt + \frac{1}{2} {αt}^{2} . . . . (i) [It is an equation of angular variables of motion corresponding to the linear variable based equations s] (s = ut + \frac{1}{2} {αt}^{2}) (i) \Rightarrow θ = 0 + \frac{1}{2} {αt}^{2} θ = \frac{{αt}^{2}}{2} = \frac{2 \times 100}{2} = 100 n = \frac{θ}{2 π} = \frac{100}{2 \times 3.14} = 15.92 \approx 16$

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NEET physics

A car of mass 1000 kg moves on a circular track of radius 20 m. If the coefficient of friction of 0.64. The maximum velocity with which the car can be moved is

11.2m/s

112m/s

$\frac{0.64 \times 20}{1000 \times 100} m / s$

$\frac{1000}{64 \times 20} m / s$

The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in $\sqrt{22}$ sec is (g=9.8 m/s)

15⁰

30⁰

60⁰

45⁰

The radius of earth is 6400 km and g = 10 m/s². In order that a body of 5 kg weight is zero at the equator, the angular speed is

$\frac{1}{800} rad / \sec$

$\frac{1}{1600} rad / \sec$

$\frac{1}{400} rad / \sec$

$\frac{1}{80} rad / \sec$

The escape velocity for the earth is 11.2 km/sec. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

56.0 km/sec

280 km/sec

112 km/sec

56 km/sec

A wheel rotates with constant acceleration of 2.0 radian/sec² .If the wheel starts from rest the number of revolution it makes in the first ten second, will be approximately

32

24

16

8

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For Daily Free Study Material Join wiredfaculty

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NEET physics

A car of mass 1000 kg moves on a circular track of radius 20 m. If the coefficient of friction of 0.64. The maximum velocity with which the car can be moved is 11.2m/s 112m/s 0.64×201000×100m/s 100064×20m/s

The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in 22 sec is (g=9.8 m/s) 150 300 600 450

The radius of earth is 6400 km and g = 10 m/s2. In order that a body of 5 kg weight is zero at the equator, the angular speed is 1800rad/sec 11600rad/sec 1400rad/sec 180rad/sec

The escape velocity for the earth is 11.2 km/sec. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is 56.0 km/sec 280 km/sec 112 km/sec 56 km/sec

A wheel rotates with constant acceleration of 2.0 radian/sec2 .If the wheel starts from rest the number of revolution it makes in the first ten second, will be approximately 32 24 16 8

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A car of mass 1000 kg moves on a circular track of radius 20 m. If the coefficient of friction of 0.64. The maximum velocity with which the car can be moved is

11.2m/s

112m/s

$\frac{0.64 \times 20}{1000 \times 100} m / s$

$\frac{1000}{64 \times 20} m / s$

The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in $\sqrt{22}$ sec is (g=9.8 m/s)

15⁰

30⁰

60⁰

45⁰

The radius of earth is 6400 km and g = 10 m/s². In order that a body of 5 kg weight is zero at the equator, the angular speed is

$\frac{1}{800} rad / \sec$

$\frac{1}{1600} rad / \sec$

$\frac{1}{400} rad / \sec$

$\frac{1}{80} rad / \sec$

The escape velocity for the earth is 11.2 km/sec. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

56.0 km/sec

280 km/sec

112 km/sec

56 km/sec

A wheel rotates with constant acceleration of 2.0 radian/sec² .If the wheel starts from rest the number of revolution it makes in the first ten second, will be approximately

32

24

16

8