A sphere is suspended by a thread of length l . The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is

Question

A sphere is suspended by a thread of length l . The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is (A) 2√g l (B) √2 g l (C) √g l (D) 4√g l

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-1qvxcdvsbbt6ke9p.jpg" /> Kinetic energy at A=(1/2)mu2 Potential energy at B=mgh For conservation of energy, (1/2)mu2=mgl u=√2 g l

Q. A sphere is suspended by a thread of length $l$ . The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is

$2 g l$

$2 g l$

$g l$

$4 g l$

Kinetic energy at $A = \frac{1}{2} m u^{2}$
Potential energy at $B = m g h$
For conservation of energy, $\frac{1}{2} m u^{2} = m g l$
$u = 2 g l$

Q. A sphere is suspended by a thread of length l . The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is

2gl​

2gl​

gl​

4gl​

Kinetic energy at A=21​mu2 Potential energy at B=mgh For conservation of energy, 21​mu2=mgl u=2gl​

Q. A sphere is suspended by a thread of length $l$ . The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is

$2 g l$

$2 g l$

$g l$

$4 g l$

Kinetic energy at $A = \frac{1}{2} m u^{2}$
Potential energy at $B = m g h$
For conservation of energy, $\frac{1}{2} m u^{2} = m g l$
$u = 2 g l$