Question

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Answer 1

SOLUTION.

HEY !

HOPE MY ANSWER FIND U IN GOOD PLACE.

IT IS DIFFICULT FOR ME TO ADD SO MANY DIFFERENT SIGN IN THIS QUESTION BUT I TRY IT.

WHY VELOCITY NOT CHANGES SOME TIME WHEN A BALL ENTERS INTO A LIQUID OF DOME DENSITY ?

We know that rate of change of velocity is known as acceleration and negative acceleration makes the velocity to loose its magnitude. And we also know that net acceleration is due to the net force so if net force acting on the body in liquid becomes zero than we can say that the velocity does not change.

OR.

FORCE OF VISCOSITY + FORCE OF BUYOANCY

WILL BALANCE ITS WEIGHT SO THAT NET FORCE IS ZERO HENCE ZERO ACCELERATION.

OHK LETS START.....

LET THE HEIGHT BE H THAN THE VELOCITY OF THE BODY AT THE SURFACE OF THE LIQUID IS √2gH

• Force of viscosity on the spherical body (V)= 6π.¶.r.v

Where v is the velocity

r is the radius

¶ is the coefficient of viscosity

• Buyoant force (B) = 4/3πr³ρg

Where ρ is the density of the liquid

• Weight (W) = 4/3πr³σg

Where σ is the density of ball

For constant velocity.

V + B = W

Put values here...

6.π.¶.r.√2gh + 4/3πr³ρg = 4/3πr³σg

Take lcm.......

9.¶.r.√2gh + 2r³ρg = 2r³σg

9.¶.r.√2gh = 2r³g(σ - ρ)

9.¶.√2gh = 2r²g(σ - ρ)

√2gh = 2r²g(σ - ρ)/9.¶

Take square on both side....

2gh = 4g²[ r²(σ - ρ)/9¶ ]²

h = 2g[ r²(σ - ρ)/9¶ ]²

OPTION 4

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#BAL

Answer 2

Answer: Option 4.

Explanation:

The answer of the user Anu24239 is well explained . This answer will show you a quicker and easier approach to your question.

$\text{Let the height from which it is dropped be h.} \\ \\ \text{From kinematics equation:} \\ \\ h = \frac{v^2}{2g} \ \ \ \ (\text{The ball is dropped hence initial velocity = 0})\implies \text{(i)} \\ \\ \text{Given that the speed of ball doesn't change on entering the liquid. } \\ \\ \text{Also in the liquid the ball moves with the attained terminal velocity.} \\ \\ \text{Therefore :}\ v = v_t \implies \text{(ii)} \\ \\ \text{We know that :}\ v_t = \frac{2r^2g(\sigma - \rho)}{9\eta}$

$\text{From (i) and (ii) we get :} \\ \\ h = \frac{v_t^2}{2g} = \frac{\big(\frac{2r^2g(\sigma - \rho)}{9\eta}\big)^2}{2g} = 2g\Big[\frac{(\sigma - \rho)r^2}{9\eta}\Big]^2$

Hope this helps.