A srone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is in the lowest position , and has a speed u. The magnitude of the change in its velocity as it reaches a position where the sting is horizontal is :a) root (u^2 - 2gL)b) root (2gL)c) root (u^2 - gL)d) root (2(u^2 - gL))please help by answering this Q and the method to do this Question

applying energy conservation

 1/2mu^2=mgl+1/2mv^2

  we get v=root(u^2-2gl)

  in vector form we can express as v vector=root(u^2-2gl)j and u vector =ui

  difference is v-u =root(u^2+v^2)=root(2(u^2-gl))

Can some1 explain this more clearly !

student-name Abhiram M V asked in PhysicsA stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in the velocity as it reaches a position where the string is horizontal is (A) u2−2gL−−−−−−−√ (B) 2gL−−−√ (C) u2−gL−−−−−−√ (D) 2(u2−gL)−−−−−−−−−√ 7 Follow 0student-name A K Daya Sir answered this2879 helpful votes in Physics, Class XI-ScienceLet the direction of revolution be counter-clockwiseInitial velocity = u i^ (at bottom)i^ ( i-cap suggests that the direction of initial velocity is towards positive x-axis)--When the string is horizontal, the body has covered a quarter of revolution, its velocity is v j^j^ (j-cap suggests that the direction of its velocity is towards positive y-axis when the string is horizontal)-Change in velocity = vj^ - ui^magnitude of change in velocity = √ v2 + u2-Using conservation of energyKinetic Energy of body when it is at bottom = 1/2 mu2When it goes to the quarter position, it does some work against gravity. So a part of its kinetic energy is spent in doing work equal to mgh and remaining kinetic energy is 1/2 mv21/2 mu2 = 1/2 mv2 + mghhere h = l1/2 mu2 = 1/2 mv2 + mglv2 - u2 = -2glv2 - u2 + 2u2 = 2u2 - 2glv2 + u2 = 2 (u2 - gl)Therefore, change in magnitude of velocity = √ v2 + u2 = √2 (u2 - gl)

Dear Student,
Please find below the solution to your problem.

applying energy conservation
1/2mu^2=mgl+1/2mv^2
we get v=root(u^2-2gl)
in vector form we can express as v vector
=root(u^2-2gl)j and
u vector =ui
difference is v-u
= root(u^2+v^2)
= root(2(u^2-gl))

Thanks and Regards