Step 1: Compound Interest Formula.
The compound interest formula is given by:
\[
A = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
where:
\(P = 15000\), \(r = 8%\), \(t = \frac{3}{2} \, \text{years}\), and \(n\) is the number of compounding periods.
Step 2: Case 1 - Compounded Semi-Annually.
For semi-annual compounding, \(n = 2\). The formula becomes:
\[
A = 15000 \left( 1 + \frac{8}{2 \times 100} \right)^{2 \times \frac{3}{2}} = 15000 \left( 1 + 0.04 \right)^{3} = 15000 \times 1.04^3
\]
Calculating \(1.04^3\), we get:
\[
A \approx 15000 \times 1.124864 = 16872.96
\]
Thus, the compound interest is:
\[
CI_{\text{semi}} = 16872.96 - 15000 = 1872.96 \approx 1873
\]
Step 3: Case 2 - Compounded Annually.
For annual compounding, \(n = 1\). The formula becomes:
\[
A = 15000 \left( 1 + \frac{8}{100} \right)^{\frac{3}{2}} = 15000 \times 1.08^{1.5}
\]
Calculating \(1.08^{1.5}\), we get:
\[
A \approx 15000 \times 1.121032 = 16815.48
\]
Thus, the compound interest is:
\[
CI_{\text{annual}} = 16815.48 - 15000 = 1815.48 \approx 1815
\]
Step 4: Difference in Interest.
The difference in compound interest is:
\[
CI_{\text{semi}} - CI_{\text{annual}} = 1873 - 1815 = 58
\]
Thus, the difference is ₹58.
Final Answer: \[ \boxed{20} \]
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: