A pendulum bob can swing along a circular path on a smooth inclined plane, as shown in figure when $m=1.2kg, \mathcal{l}=0.75m, \theta ={37}^0.$ At the lowest point of the circle the tension in the string is $T=110N.$ Choose the correct option(s)

 

From Newton’s second law, at the lowest point, we obtain $T_A-mg\sin \theta =\frac{m{v}_0^2}{l}$

$v_0^2=\frac{T_{A}l}{m}-gl\sin \theta =\frac{\left(110\right)\left(0.75\right)}{\left(1.2\right)}-\left(9.8\right)\left(0.75\right)\left(sin{37}^0\right)$

$v_0^2=64.34$

$v_0=8.02m/s$

At the highest point $T_B+mg\sin \theta =\frac{m{v}^2}{l}$

From energy conservation between position A and position B 

$K.E_A+U_A=K.E_B+U_B$

$\frac{1}{2}m{v}_0^2+0=\frac{1}{2}m{v}^2+2mgl\sin \theta$

$v^2=v_0^2-4gl\sin \theta =\left(64.34\right)-4\times \left(9.8\right)\left(0.75\right)\left(\sin {37}^0\right)=46.7$

$v=6.8m/s$

At the highest point $T_B+mg\sin \theta =\frac{m{v}^2}{l}$

$\Rightarrow T_B=\frac{m{v}^2}{l}-mg\sin \theta =\frac{\left(1.2\right)\left(46.7\right)}{0.75}-\left(1.2\right)\left(9.8\right)\left(\sin {37}^0\right)=67.66N$