Question

$\bf{\fbox\red {Physics}}$

Explain briefly.​

Answer 1

Answer:

$(2) \mathrm{\ 10\sqrt{10}\ m/s^2}$ is the answer

Step-by-step explanation:

Given :

At B, tension = 0

g = 10 m/s²

To find :

The acceleration of bob at point A

Solution :

At B, the forces acting on bob be centripetal force, weight and tension

$\implies \mathrm{mg + T = \dfrac{mv^2}{l}}$

As T = 0,

$\implies \mathrm{mg= \dfrac{mv^2}{l}}$

$\implies \mathrm{\not mg= \dfrac{\not mv^2}{l}}$

$\implies \mathrm{v^2 = gl}$

$\implies \mathrm{v = \sqrt{gl}}$ ----[1]

At A, Energy conservation,

PE₁ + KE₁ = PE₂ + KE₂

Let $\mathrm{v_A}$ be the velocity at A

$\mathrm{\dfrac{1}{2}mv^2 + mgl = \dfrac{1}{2}mv_A^2}$

From [1],

$\implies \mathrm{\dfrac{1}{2}mgl + mgl = \dfrac{1}{2}mv_A^2}$

$\implies \mathrm{\dfrac{3}{\not2}\not mgl = \dfrac{1}{\not2}\not mv_A^2}$

$\implies \mathrm{v_A^2 = 3gl}$

$\implies \mathrm{v_A = \sqrt{3gl}}$

We know that,

$\boxed{\mathrm{Centrapetal\ force = a_c = \dfrac{v^2}{r}}}$

So,

$\implies \mathrm{a_c = \dfrac{3g \not l}{\not l}}$

$\implies \mathrm{a_c = 3g}$

We know that, Resultant acceleration(a)

$\boxed{\mathrm{a = \sqrt{a_t^2+a^2_c}}}$

At point a tangential acceleration is g as it is tangent at that point downward. Thus, $\mathrm{a_t = g}$

Applying the formula,

$\implies \mathrm{a = \sqrt{g^2+(3g)^2}}$

$\implies \mathrm{a = \sqrt{g^2+9g^2}}$

$\implies \mathrm{a = \sqrt{10g^2}}$

$\implies \mathrm{a = g\sqrt{10}}$

As g = 10 m/s²,

$\implies \mathrm{a = 10\sqrt{10}\ m/s^2}$

Thus, 10√10 is the answer

Hope it helps!!