Laws Of Motion Question 1
Question 1 - 24 January - Shift 1
As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if $g=$ $10 m / s^{2}$ )
(1)$(4 \sqrt{3}+1) N$
(2) $4(\sqrt{3}+1) N$
(3) $4(\sqrt{3}-1) N$
(4) $(4 \sqrt{3}-1) N$
Show Answer
Answer: (2)
Solution:
Formula: Second Law Of Motion
$4 g \sin 60^{\circ}-T=4 a$
$4 g \sin 60^{\circ}$
$g \sin 30^{\circ}$
$T-g \sin 30^{\circ}=a$
Solving (1) and (2) we get.
$20 \sqrt{3}-T=4 T-20$
$T=4(\sqrt{3}+1) N$
