The operation used to obtain a scalar from two vectors is ______
a) Cross product
b) Dot product
c) Simple product
d) Complex product

Answer

A, B

Explanation

Let us consider a small element which makes angle `dtheta` at the centre.
`:. dm=rho (m/L)Rdtheta`
a. Gravitational potential energy of dm with respect to centre of the sphere
`(dm)gRcostheta`
`=((mg)/L)R^2costheta dtheta

`:. total G.P.E. =int_0^(L/R) mg R^2/L costheta dtheta`
`=(L/R)`
`=(mR^2g)/L[sintheta][theta=L/R]`
`=(mR^2g)/Lsin(L/R)`
b. when the chain is released from rest and slides down thrugh an angle theta, the K.E. of the chain is given by K.E. =change is potential energy.
`(mR^2g)/L sin (L/R)-int(gR^2)/L costheta dtheta`
`=(mR^2g)/L[sin(L/R)+sintehta-sin{theta+(L/R)}]`
c. Since `K.E. =1/2 mv^2`
`=(mR^2g)/L[sin(L/R)]`
taking derivative of both sides with respect to t,
`(1/2)xx2xxx(dv)/(dt)`
`=(R^2g)/L[costheta-(dtheta)/dt)-cos(theta+L/R)(dtheta)/(dt)]`
`:. [R-(dtheta)/(dt) (dv)/dt`
`=(R^2g)/Lxx(dtheta)/(dt)[costhet-cos(theta+L/R)]`
`(because`v=Romega=R (dtheta)/(dt))`
`:. (dv)/(dt)=(Rg)/L[costheta-cos(theta+L/R)]`
When the chain strrts sliding down,
`theta=0`
`:. (dv)/(dt)=(Rg)/L[1-cos(L/R)]`