Question

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and the string vibrates forming 8 loops. When the stone is immersed in water 10 loops are formed. The specific gravity of the stone is close to
 
A)  1.8
B)  4.2
C)  2.8
D)  3.2​

Answer 1

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)  

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$      ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength   .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$   =   f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$   =   f × $\lambda_1$     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$      ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g    

from equation 2

f × $\lambda_2$ = f  × $\frac{1}{5}$  

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$     .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

so

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8