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Question

A wire suspended from one end carries a sphere at its other end. The elongation in the wire reduces from $$2 mm$$ to $$1.6 mm$$ on completely immersing the sphere in water. The density of the material of the sphere is


A
3200 kg/m3
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B
800 kg/m3
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C
1250 kg/m3
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D
5000 kg/m3
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Solution

The correct option is D 5000 kg/m$$^{3}$$
$$ Y = \dfrac{F/A}{\triangle L/L}$$
$$ Y = \dfrac{F_{1}/A}{2/L} = \dfrac{F_{2}/A}{1.6/L}$$
$$\dfrac { 4  }{ 5 } F_{1} = F_{2}$$
Bouyant  force  $$= F_{1} -F_{2} = \dfrac{F_{1}}{5}$$
$$ F_{1} = \rho_{s} V_{s} g$$
Bouyant  force  $$=  \rho_{w} V_{s} g = \dfrac{F_{1}}{5}$$ 
$$ \Rightarrow  5\rho_{w} V_{s} g = \rho_{s} V_{s} g$$
$$ \Rightarrow \rho_{s} = 5\rho_{w}$$
$$ \Rightarrow \rho_{s} = 5000 kg / m^{3}   (\because \rho_{w} = 1000  kg/m^{3})$$
So,  density  of  material  is  $$5000 kg / m^{3}$$

Physics

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