Question

A sphere of mass m is held between two smooth inclined walls. For $\sin {37}^{\circ }=\frac{3}{5},$ the normal reaction of the wall (2) is equal to

• A. $\frac{16mg}{25}$
• B. $\frac{25mg}{21}$
• C. $\frac{39mg}{21}$
• D. $mg$

Answer 1

Answer: (D)

$mg$

In horizontal duration

$T_2\sin (2\times {37}^{\circ })=T_1\sin {37}^{\circ }$

$2T_2\sin {37}^{\circ }\cos {37}^{\circ }=T_1\sin {37}^{\circ }$

$2\times T_2\times \frac{4}{5}=T_1$

$T_1=\frac{8T_2}{5}$

Now in vertical direction

$T_1\cos {37}^{\circ }=mg+T_2\cos (2\times {37}^{\circ })$

$\frac{8}{5}{T}_2\times \frac{4}{5}=mg+T_2\times (-1+2{\cos }^2{37}^{\circ })$

$\frac{32}{25}{T}_2=mg+T_2(-1+2\times (\frac{4}{5}{)}^2)$

$=\frac{32}{25}{T}_2=mg+T_2(\frac{32}{25}-1)$

$\frac{32}{25}{T}_2=mg+\frac{7T_2}{25}$

$\frac{32-7}{25}{T}_2=mg$

$\frac{25}{25}{T}_2=mg$

$T_2=mg$