Question

An object of density 12g/ cm cube is weighed with brass weights of density 8 g/ cm cube by a physical balance.if the density of air is 12× 10 power minus 3 g/ cm cube then the % error in weighing will be

Answer 1

ise the formula,
$\Delta{W}=\frac{\rho_{air}}{\rho_O}-\frac{\rho_{air}}{\rho_{brass}}$
where ∆W is fractional error in weighting, $\rho_{air}$ density of air , $\rho_O$ is density of object and $\rho_{brass}$ is density of brass.

now,
∆W = 0.012/1.2 - 0.012/8
= 0.012[5/6 - 1/8]
= 0.012 × (40 - 6)/48
= 0.012 × 34/48
= 0.034/4
= 0.017/2
= 0.0085

so, % error = 0.0085 × 100 = 0.85 %

Answer 2

Answer:

0.005%

Explanation:

Weight of cube in vacuum:- mg

Volume of the cube :- m/12

Volume of brass:-m/8

Buoyancy of in cube :- m/12 × 12×10¯³ × g

Buoyancy of in brass :- m/8 × 12×10¯3 ×g

Error % = (m/12 × 12×10¯3×g - m/8 × 12×10¯3 ×g) /mg × 100

= m*12*10¯3*g(1/12-1/8)/ mg *100

After solving u will get = 0.005%