When switch is shifted from 1 to 2 then find charge flown in circuit and heat generated.

Explanation

Initially `q_(1)/C1-5+10=0`
`rArr q_(1)=10muF`
Finally `rArr q_(2)=10 xx 2muC=20muC`

So charge Flow `(Q)=(q_(2)-q_(1))=10muC`
Work done by battery `=QV=10 xx 10xx 10=100muJ`

So, Heat `=[W.D_(B)-"{final initial} energy of capacitor]"`
`=[100-{1/2 xx 2xx (0)^(2)-1/2xx 2 xx 5}]muJ`
`=[100-{100-25}muJ=25muJ`