2018-10-01_Physics_For_You

17. (i) A balloon filled with helium goes on rising
    in air as long as the weight of the air displaced by it
    (i.e., upthrust) is greater than the weight of filled
    balloon. We know that the density of air decreases with
    height. Therefore, the balloon halts after attaining a
    height at which density of air is such that the weight
    of air displaced just equals the weight of helium filled
    balloon.
    (ii) Water exerts much more upthrust on the limbs
    of man than air. So the net weight of limbs in water is
    much less than that in air. Hence the force required by
    a man to raise his limbs immersed in water is smaller
    than the force for the same movement in air.


18. In the shown figure, v^22 = v^20 + 2gh
    and A 1 v 0 = A 2 v 2

Solving,

A

A

v

vgh

2

1

0

0

(^22)

=

+

A

A

A

A

v

vgh

2

1

2

2

0

0

(^222)

==

+

4 v 02 = v 02 + 2gh Ÿ h

v

g

=

3

2

0

2

19. (a) Young's modulus,
    Y Mgl
    rl

Mgl

D

l

Mgl

Dl

==









=

π

π

(^22) π 2
2

4

.

.

∆.

∆

∆

where D is the diameter of the wire.

Elongation, ∆∆l

Mgl

DY

ie l

D

=∝

(^41)
π^22 ..,
Clearly, if the diameter is doubled, the elongation will
become one fourth.
(b) Load, Mg
DlY
l
= ie Mg D

⋅⋅

∝

π^22

4

∆

..,

Clearly, if the diameter is doubled, the wire can support
4 times the original load.

20. (i) Effective value of acceleration due to gravity
    becomes (g + a 0 ).

Required velocity of efflux, vg=+ 2 ()ah 0
(ii) Equating the rate of flow, v 1 A 1 = v 2 A 2
But vg 11 == 22 yA,,Lv^222 =×gy 4 ,AR=π^2

∴× 22 gy Lg^22 =× 4 yR×π (using Torricelli's law)

or LRor R

22 L

2

2

==π

π

A 1 , v 0

A 2 , v 2

h

21. (a) Velocity of efflux, vg= 2 h

h

H–h

v

H

s

Time taken by the liquid to touch the ground, i.e., to

travel a vertical distance (H – h) is given by,

()

()

Hh gt t

Hh

g

−= =

1 −

2

2 2

or

Thus, s = vt =^2

2

gh

Hh

g

×

()−

or sh=−2(Hh)

(b) For s to be maximum, ds

dh

= 0

or

d

dh

[( 20 hH−=h)]

or 2 × (1/2) (Hh – h^2 )–1/2 (H – 2h)= 0

or Hh

Hh h

−

−

=

2

0

2

or h=H

2

22. (i) Let L be the greatest length of aluminium wire
    that can hang without breaking.
    Mass of the wire, M = ALρ

Stress = Mg

A

AL g

A

=

()ρ = Lρg

As breaking stress = 7.5 × 10^7 N m–2,

Lρg = 7.5 × 10^7 N m–2

or L

g

=

75. × 107

ρ

=

×

××

7510

27 10 98

7

3

.

..

= 2.8 × 10^3 m = 2.8 km

(ii) For a material, shear modulus is smaller than the

Young's modulus. This shows that it is easier to slide

layers of atoms of solid over one another than to pull

them apart or to squeeze them close together.

23. Density of sphere, V = 0.5 × 10^3 kg m–3
    (i) Here ρ = 10^3 kg m–3, g = 10 m s–2,
    a = 2 m s–2, m = 2 kg

Volume of sphere, V=m

σ

Weight of sphere in accelerating

medium = VV(g + a)

Upthrust on sphere due to liquid

= ρV(g + a) VV(g + a)

ρV(g + a)

T