TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, mass number, M1=M
Initial volume, V1=V
Final mass number, M2=2M
As we know that,
‌‌R=R0A1∕3
⇒‌‌R∝A1∕3 . . . (i)
‌ where, ‌R0=‌ radius of nucleus ‌
‌ and ‌‌‌A=‌ mass number. ‌
Since, volume V=‌

4

3

πR3 . . . (ii)
⇒‌‌R∝(V)1∕3
Now, from Eqs. (i) and (ii), we get
⇒‌R∝A1∕3∝V1∕3
⇒‌‌

A1

A2

=‌

V1

V2

⇒‌‌

M1

M2

=‌

V1

V2

‌

M

2M

‌=‌

V

V2

⇒V2=2V