Question

4th Question.Please.

Answer 1

Hey mate!!

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your answer is option (a)

FIRST INCREASES THEN DECREASES.

EXPLANATION:-

Here, HOLLOW sphere acts as simple pendulum.

so, it's Time period will be

T = 2π √ l/mgr

where,

l = inertia of rotation of sphere

l=mK^2

so,

T = 2π√ mk^2/mgd

T = 2π√ k^2/gd

now,

in question it is given, as water flow out the distance of centre of axis from the C.O.M increases so, redius of gyration increases in square.

TIME PERIOD OF (PENDULUM)HOLLOW SPHERE INCREASES.

now,

when after some time the rate of decrease in d increases as c.o.m shifts to heavier mass but redius of gyration (K) remains constant.

overall value of k^2/d decreases.

Time period of pendulum decreases.

So, your answer is option a ....1st increases then decreases.

hope it helps!!

#phoenix