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# A particle P is sliding down a frictionless hemisphere bowl. it passes the point A at t=0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t=0 along horizontal spring AB with the speed v. Friction between bead and the string may be neglected. Let t(p) and t(q) be the respective times taken by P and Q tp reach the point B then A. t(p)=t(q) B. t(p)>t(q) C. t(p) ## 4 Answers

10 years ago

the mass of the particle P and the bead Q is the same and it is also given that they have the same velocity.

the object is a hemispherical bowl .

Let D1 be distance travelled by the particle P with mass m and velocity v.

D1 = vt

and D2 be the distance travelled by the bead Q with mass m and velocity v

since D2 is actually the displacement of Particle P and distance of bead B

we know displacement <= distance

displacement of P < distance travelled by P

==> distance travelled by bead Q < distance travelled by P

since velocity is the same.

time taken to travel a smaller distance is less than time taken to travel a larger distance.

therefore time taken by Particle P > time taken by bead Q

option B t(P) > t(Q) is the final answer

please approve if the answer is correct

10 years ago

SINCE THE VELOCITY COMPONENT IN THE HORIZONTAL DIRECTION IS SAME FOR BOTH THE BEADS AND THERE IS NO FRICTION IN BOTH THE PATHS . THEREFORE T(P)=T(Q).

one year ago
At A the horizontal speeds of both the masses are the same. The velocity of Q remains the same in horizontal as no force is acting in the horizontal direction. But in case of P as shown at any intermediate position, the horizontal velocity first increases (due to force Nsintheta), reaches a max value at O and then decreases. Thus it always remains greater than v. Therefore tp

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