Case I : Given
P = 1 atm
w = 12 g
T = (t+273)K
V = V litre
Case ii :
T = (t+283)K
P = 1 + $\large\frac{10}{100}$ = 1.1 atm
w = 12 g
V = V l
Using gas equation
Case I : $1\times V = \large\frac{12}{m}\times R(t + 273)$ ----(i)
Case II : $1.1\times V = \large\frac{12}{m}\times R(t + 283)$ ----(ii)
By equation (i) and (ii)
$\large\frac{1.1}{1} = \large\frac{t + 283}{t + 273}$
$\therefore 1.1t + 300.3 = t + 283$
0.1t = -17.3
$t = -173^{\large\circ}C$ = 100K
Also from case I (Since m = 120)
$1\times V = \large\frac{12}{120}\times0.082\times 100$
V = 0.82 l
Hence answer is (A)