Question

A metal cube of 2cm edge weighs 0.56 N in air taking g=10 Ms , Calculate , it's apparent weight when immersed in spirit of density 0.85gcm , and the density of metal of which it is made
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Answer 1

Explanation:

CUBE = 2cm = 2/100 m = 2*10^-2m

Given, length of cube, l = 2*10^-2m

Volume V = l^3 = 8*10^-6m

Let's, density = p (sorry, it's raw not P I'm taking p)

then, Mass = p*v

M = p*8*10^-6*10

= p*8*10^-3

therefore, 0.56 = p*8*10^-6*10

= p*7*10^3 kg m^-3

Therefore, Density of material p= 7*10^3 kg m^-3

when it is submerged in a liquid with density P(L), it's apparent weight reduces due to Buoyant Force on it

Buoyant force N= P(L) Vg

Given, P(L) = 0.85g cm^-3

= 0.85*10^3 kg m^-3

N = 0.85 * 10^3 * 8 * 10^-6 * 10

= 68 * 10^-3 N

= 0.068 N

Apparent weight W1 = W-N

= (0.56-0.068) N

= 0.492 N