What is a scalar?
a) A quantity with only magnitude
b) A quantity with only direction
c) A quantity with both magnitude and direction
d) A quantity without magnitude

Answer

A, B, C

Explanation

a. When the bob has an initial height less than the peg and the released from rest ure let body travels from A to B since, total energy at A
`=Total energy at B,
`:.(KE)_A+(PE)_A=(KE)_B+(PE)_B`
`rarr(PE)_=(PE)_A`
because `(KE)A=(KE)_B=0`
So the maximum height reached by the bob is equal to initial height.
b. When the pendulum is released with `theta ==90^0 and x=l/2, ure the path of the particle is shown in ure.

At point C the string wil becomke slack and so the particle wil start making projectile motion.
Apply law of conservation.
`(1/2)mv_c^2-0=mg(L/2)(1-cosalpha)`
[because distance between A and C ]
In the vertical direction is
`L/2=(1-cosalpha)`
`rarr V_c^4=gL(1-cosalpha)`............i
Again, from the feebody diagram ure
`(mv_c^2)/(L/2)=mgcos alpha`......ii
`[becuse T_C=0]`
From equation i and equation ii
`gL(1-cosalpha)=(gL)/2cosalpha`
`rarr 1-cosalpha=1/2cosalpha`
`rarr 3/2 cosalpha=1`
`rarr cosalpha=(2/3)`......iii
to find highest positon `C_1 ` before the string becomes slack.
`BF=L/2+L/2 costheta`
`=L/2+L/2xx2/3`
`=L(1/2+1/3)`
So, `BF=((5L)/6)`
c. If the principle has to complete a vertical circle at the point C
`(mv_c^2)/(L-x)=mg`.......i
Again applying energy principle between A and C
`(1/2)mv_c^2-0=mg(OC)`
`rarr (1/2)mv_c^2=mg(L-2(l-x))`
`=mg(2x-L)`
`rarr v_c^2=2g(2x-L)` .......ii
From equation i and ii
`g(L-x)=2g(2x-L)`
`rarr L-x=4x-2L`
`rarr 5x=3L`
`:. x/L=3/5=0.6`
So, the rates `(x/L)` should be 0.6.