A block of mass m and height h is hanging in a liquid kept in a container upto height 5h as shown in the figure. Density of block is two times of density of liquid. A very small hole is made at the bottom of container. Tension in the string varies with height x of the liquid surface from bottom as

$T = a (b - \frac{x}{h}) f o r 2 h \leq x \leq 3 h$

What is the value of a?

2 mg

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\frac{m g}{2}

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\frac{m g}{4}

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Solution

The correct option is C $\frac{m g}{2}$
At $x = 3 h, T = m g - F_{B} = m g - \frac{ρ}{2} V g = \frac{m g}{2}$

At $x = 2 h, T = m g$

$∴ \frac{m g}{2} = a (b - \frac{3 h}{h}) a n d m g = a (b - \frac{2 h}{h}) \Rightarrow a = \frac{m g}{2} a n d b = 4$

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