A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is

Question

A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is (A) (√2gR/g cos θ) (B) 2√gR. (cos θ/g) (C) 2√(R/g) (D) (gR/√g cos θ)

Tardigrade Admin · Accepted Answer

Acceleration of body along AB is g cos θ. Distance travelled in time t sec is AB =(1/2) (g cos θ)t2 From Δ;ABC, AB = 2R cos θ Thus, 2 R cos θ (1/2) g cos θ t2 ⇒ t2=(4 R/g) ⇒ t=2√(R/g)

Q. A frictionless wire $A B$ is fixed on a sphere of radius $R$ . A very small spherical ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is

$\frac{2 g R}{g cos θ}$

$2 g R . \frac{cos θ}{g}$

$2 \frac{R}{g}$

$\frac{g R}{g cos θ}$

Acceleration of body along $A B$ is $g cos θ$ . Distance travelled in time $t$ sec is
$A B = \frac{1}{2}$ $(g cos θ) t^{2}$

From $Δ; A BC, A B = 2 R cos θ$
Thus, $2 R cos θ \frac{1}{2} g cos θ t^{2}$
$\Rightarrow t^{2} = \frac{4 R}{g} \Rightarrow t = 2 \frac{R}{g}$

Q. A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is

gcosθ2gR​​

2gR​.gcosθ​

2gR​​

gcosθ​gR​