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(c) Rs 961

Here, $A=P \times\left(1+\frac{R}{100}\right)^{2} \times\left(1+\frac{\frac{1}{4} R}{100}\right)$

$=$ Rs. $4000 \times\left(1+\frac{10}{100}\right)^{2} \times\left(1+\frac{\frac{1}{4} \times 10}{100}\right)$

$=$ Rs. $4000 \times\left(\frac{100+10}{100}\right)^{2} \times\left(\frac{40+1}{40}\right)$

$=$ Rs. $4000 \times\left(\frac{110}{100}\right)^{2} \times\left(\frac{41}{40}\right)$

$=$ Rs. $4000 \times\left(\frac{11}{10}\right) \times\left(\frac{11}{10}\right) \times\left(\frac{41}{40}\right)$

$=$ Rs. $(11 \times 11 \times 41)$

$=$ Rs. 4961

$\therefore$ Compound interest $=$ amount $-$ principal $=$ Rs $(4961-4000)=$ Rs 961