A) 104 g
B) 58 g
C) 208 g
D) 54 g
Correct Answer: A
Solution :
\[mg+Mg={{\rho }_{\omega }}Vg\] |
\[m'g+Mg=\frac{{{\rho }_{\omega }}}{1+\gamma \Delta T}Vg\] |
\[m-m'=\rho \left( \omega -\frac{{{\rho }_{\omega }}}{1+\gamma \Delta T} \right)g\left( \frac{{{\rho }_{\omega }}\gamma \Delta T}{1+\gamma \Delta T} \right){{V}_{g}}\] |
\[52g={{\rho }_{\omega }}Vg\] |
\[2-m'\approx \frac{52\times {{10}^{-4}}\times 20}{1}=104\,gm.\] |
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