2018-10-01_Physics_For_You

12. The bubble will
    separate from the ring
    when
    2 pb × 2V sin q = ρAv^2

or 4 πσb ρπ^22

b

R

×=××bv

or R
v

=

4

2

σ

ρ

13. The lead slab is fixed and force is applied parallel
    to narrow face as shown in figure. The area of the face
    parallel to which force is applied is
    A = 50 cm × 10 cm = 0.5 m × 0.1 m = 0.05 m^2

F

50 cm qq

Therefore, the stress applied =^9010
005

4

2

.

.

× N

m
= 1.80 × 10^6 N m–2

We know that shearing strain, ∆x
LG

=Stress

Therefore, the displacement, 'x =

Stress×L

G

∆x=

××

×

−

−

18 10 05

56 10

62

92

..

.

Nm m

Nm

= 1.6 × 10–4 m

'x = 0.16 mm

14. Torricelli's theorem : It states that velocity of liquid
    flowing out of a narrow hole (i.e., velocity of efflux) at
    a depth h below the free surface of liquid in a vessel is

vg= 2 h. Let us consider a tank containing an ideal
liquid and having a hole at O as shown in figure. Further,
let the flow of the liquid be streamlined.

F O v

vA = 0

PA = P, hA = h, vA = 0

PO = P, hO = 0, vO = v

A

s

h

h'

Let h = height of the liquid surface above the hole,
h' = height of the hole from the ground,
s = horizontal distance travelled by the liquid,

v = speed with which the liquid comes out of O,

ρ = density of the liquid,

P = atmospheric pressure.

According to Bernoulli's theorem,

total energy of the liquid at A = total energy of the

liquid at O,

i.e., PgAA++ρρhvAO=+PgρρhvOO+

1

2

1

2

22

or Pg++ρρhP 00 =++^1 v

2

2

(as pressure at A = pressure at O = atmospheric pressure

= P. At A, vA = 0 and at O, hO = 0)

Thus, gh= v

1

2

(^2) or vg= 2 h
v is called the speed of efflux of the liquid.

15. (i) The gravity on moon is about one-sixth of that
    on the earth. But gravity has equal effect both on weight
    of the body and the upthrust. So equilibrium of the
    floating body is not affected. On the earth, weight of the
    floating body is balanced by upthrust due to both water
    and air.
    ? W = mg = Vwρwg + Vaρa g
    or m = Vwρw + Vaρa ...(i)
    But the moon has no atmosphere. So
    W = mg = V wcρwg
    or m = V cwρw ...(ii)

From eqn. (i) and (ii), we note that V wc = Vw +

Vaa

w

ρ

ρ

Clearly, V cw > Vw

That is, the volume of ball immersed in water on the

moon is greater than that on earth. Hence ball will sink

slightly more in water when taken to the moon.

(ii) [F] = [MLT –2]

[6pKrv] = [ML–1T –1] [L] [LT –1] = [MLT –2]

? Dimensions of LHS = Dimensions of RHS.

Hence the given formula for viscous force F is

dimensionally correct.

16. (a) Wire of material A with larger plastic region is
    more ductile.

(b) Young's modulus is

Stress

Strain

? YA > YB

(c) For given strain, larger stress is required for A than

that for B.

? A is stronger than B.

(d) Material with smaller plastic region is more brittle,

therefore B is more brittle than A.