This document contains two physics questions and their solutions. The first question asks about the force of friction on a 10kg block with a coefficient of friction of 0.6 when a 50N force is applied. The solution shows that the frictional force is 60N, but since the applied force is only 50N, the block will not move. The second question asks about the number of beats that would be heard from three tuning forks with frequencies of 400Hz, 401Hz, and 402Hz sounded together. The solution states that there would be two beats per second, or a total of 6 beats.

4. 1 Which of the following quantities have
unit, but not the dimensions.
(a) dielectric constant
(b) coefficient of viscosity
(c) plane angle
(d) refractive index

5. 2. A block of mass 10kg is resting on a
horizontal surface. A force of 50N is
applied on it. If coefficient of friction
between the block and the surface is
0.6. what will be the force of friction?
(g = 10 m/s2)
(a) Zero
(b) 50 N
(c) 60 N
(d) 40 N

6. CORRECT OPTION IS
(b) 50 N
fs = mgμ
fs = 10×10×0.6
fs = 60N
But applied force is 50N.
Hence, the block will not move.

8. 1. Three tuning forks of frequencies
400 H2, 401 Hz and 402 Hz are
sounded together. How many beats
will be heard?
(a) 3
(b) 5
(c) 6
(d)1

9. 2. Three tuning forks of frequencies
400Hz, 401Hz and 402H2 are
sounded together. How many beats
will be produced?
(a) 3
(b) 5
(c) 6
(d)1

10. Solution:

11. g
R cos
R cos
g
R cos
2
g
1 g
2 Rcos
3. A hemispherical bowl of radius R is
revolved around its vertical axis. A Light
weight sphere in it also revolves in
it, without sliping. The radius of the
hemispherical passing through the sphere
makes angle θ with the axis of rotation. The
angular velocity of the sphere is:
(a) (b)
(c) (d)

12. Solution:
Motion of the sphere is like motion of
conical pendulum.
Hence R cos
T 2
g
2
but T
2 R cos
2
g
g
R cos

14. 1.Charge q moves around charge Q in
circular path. The work done by
charge Q on q is
(a) (b)
(c) Zero (d)
2
0
1 Qq
4 k r
0
1 q
4 k r
2
0
1 Qq
2 r
4 k r

15. Solution: Charge q
is moving over an
equipotential
surface Hence,
work done is zero.

16. 2. Time taken by radio waves to go
come back from communication
satellite is nearly
(a) 1s (b)
(c) (d) 4s
1
s
4
1
s
8

17. Height of communication satellite
above the surface of the earth is
36000 km.
∴ Distance travelled by radio waves
= 72000 km
Velocity of radio waves = 3 × 108 m/s
8
6
6
72000km
time
3 10 m / s
72 10 1
time
300 10 4

18. Full Portion Tests
Time Table
Time Allotment
Mistakes
Confidence