A wire loaded by a weight of density `7.8 g//c c` is found to be of length 100 cm. On immersing the weight in water. The length decrease by 0.20 cm find the original length of the wire.

Answer

98.44 cm

Explanation

Let l be the length and A be the area of cross- section of the wire. Let V be the volume of the weight attached to the wire. The `F =Mg = V rho g = Vxx7.8 xx 980 dyn e ` Extension in wire, `Delta l = (100 - l)cm.` Now `Y = (F)/(A) = (l)/(Delta l) = ((Vxx7.8 xx980)xxl)/(Axx(100 - l)) `...... (i) When weight is immersed in water, then apperent weight, F = true weight - up thrust of water ` = V rho g - V rho_w g =V(rho - rho_w)g`
` =V(7.8 - 1)xx980 = Vxx6.8 xx 980 dyn e` Length of wire, `Delta l' (99.80 - l)`
Now `Y = (F')/(A)xx(l)/(Delta l') = ((Vxx6.8 xx 980)xl)/(Axx(99.80 -1)) `..... (ii)
Equating (i) and (ii) we have
`(Vxx7.8 xx980 xxl)/(A(100 -l)) = (Vxx6.8xx980xxl)/(A xx (99.80 -))`
or `7644 (99.80 - l) = 6664 (100 -l)` On solving, l = 98.44 cm