Explanation:

D
Distance from wall \(\left(x_{1}\right)=\frac{L}{2}\)
Distance from wall \(\left(\mathrm{x}_{2}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}=\mathrm{L}\)
Distance from wall \(\left(\mathrm{x}_{3}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{4}=\frac{5}{4} \mathrm{~L}\)
Now,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}+M_{3} x_{3}}{M_{1}+M_{2}+M_{3}}\)
\(X_{c m}=\frac{M \times \frac{L}{2}+M \times L+M \times \frac{5 L}{4}}{3 M}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{\frac{11}{4} \mathrm{ML}}{3 \mathrm{M}}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{11}{12} \mathrm{~L}\)

Explanation:

A Given,
\(\mathrm{m}_{1}=4 \mathrm{~kg}\)
\(\mathrm{~m}_{2}=5 \mathrm{~kg}\)
\(\mathrm{v}_{1}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s} \text { (x direction) }\)
\(\mathrm{v}_{2}=3 \hat{\mathrm{j}} / \mathrm{s} \text { (y direction) }\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{4 \times 5 \hat{\mathrm{i}}+5 \times 3 \hat{\mathrm{j}}}{9}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{20}{9} \hat{\mathrm{i}}+\frac{15}{9} \hat{\mathrm{j}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\left(\frac{20}{9}\right)^{2}+\left(\frac{15}{9}\right)^{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{400+225}{81}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{25}{9} \mathrm{~m} / \mathrm{s}\)

Explanation:

B We know that,
\(\sigma=\frac{\text { mass }}{\text { area }}\)
Mass of the bigger disc-
\(\mathrm{M}_{1}=\sigma \times \pi \times(2 \mathrm{R})^{2}\)
\(\mathrm{M}_{1}=4 \pi \mathrm{R}^{2} \sigma\)
Mass of the smaller disc-
\(\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}^{2}\)

Centre of the mass of cutting disc,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}}{M_{1}+M_{2}}\)
\(X_{c m}=\frac{\left[4 \pi R^{2} \sigma \times 0\right]+\left[\left(-\sigma \times \pi R^{2}\right) \times R\right]}{\left[4 \pi R^{2} \sigma+\left(-\sigma \pi R^{2}\right)\right]}=-\frac{R}{3}\)
\(\alpha R=-\frac{R}{3} \quad\left(\because x_{c m}=\alpha R\right)\)
\(|\alpha|=\left|-\frac{1}{3}\right|\)
\(\alpha=\frac{1}{3}\)

Explanation:

Explanation:

D
Distance from wall \(\left(x_{1}\right)=\frac{L}{2}\)
Distance from wall \(\left(\mathrm{x}_{2}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}=\mathrm{L}\)
Distance from wall \(\left(\mathrm{x}_{3}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{4}=\frac{5}{4} \mathrm{~L}\)
Now,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}+M_{3} x_{3}}{M_{1}+M_{2}+M_{3}}\)
\(X_{c m}=\frac{M \times \frac{L}{2}+M \times L+M \times \frac{5 L}{4}}{3 M}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{\frac{11}{4} \mathrm{ML}}{3 \mathrm{M}}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{11}{12} \mathrm{~L}\)

Explanation:

A Given,
\(\mathrm{m}_{1}=4 \mathrm{~kg}\)
\(\mathrm{~m}_{2}=5 \mathrm{~kg}\)
\(\mathrm{v}_{1}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s} \text { (x direction) }\)
\(\mathrm{v}_{2}=3 \hat{\mathrm{j}} / \mathrm{s} \text { (y direction) }\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{4 \times 5 \hat{\mathrm{i}}+5 \times 3 \hat{\mathrm{j}}}{9}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{20}{9} \hat{\mathrm{i}}+\frac{15}{9} \hat{\mathrm{j}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\left(\frac{20}{9}\right)^{2}+\left(\frac{15}{9}\right)^{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{400+225}{81}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{25}{9} \mathrm{~m} / \mathrm{s}\)

Explanation:

B We know that,
\(\sigma=\frac{\text { mass }}{\text { area }}\)
Mass of the bigger disc-
\(\mathrm{M}_{1}=\sigma \times \pi \times(2 \mathrm{R})^{2}\)
\(\mathrm{M}_{1}=4 \pi \mathrm{R}^{2} \sigma\)
Mass of the smaller disc-
\(\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}^{2}\)

Centre of the mass of cutting disc,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}}{M_{1}+M_{2}}\)
\(X_{c m}=\frac{\left[4 \pi R^{2} \sigma \times 0\right]+\left[\left(-\sigma \times \pi R^{2}\right) \times R\right]}{\left[4 \pi R^{2} \sigma+\left(-\sigma \pi R^{2}\right)\right]}=-\frac{R}{3}\)
\(\alpha R=-\frac{R}{3} \quad\left(\because x_{c m}=\alpha R\right)\)
\(|\alpha|=\left|-\frac{1}{3}\right|\)
\(\alpha=\frac{1}{3}\)

Explanation:

Explanation:

D
Distance from wall \(\left(x_{1}\right)=\frac{L}{2}\)
Distance from wall \(\left(\mathrm{x}_{2}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}=\mathrm{L}\)
Distance from wall \(\left(\mathrm{x}_{3}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{4}=\frac{5}{4} \mathrm{~L}\)
Now,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}+M_{3} x_{3}}{M_{1}+M_{2}+M_{3}}\)
\(X_{c m}=\frac{M \times \frac{L}{2}+M \times L+M \times \frac{5 L}{4}}{3 M}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{\frac{11}{4} \mathrm{ML}}{3 \mathrm{M}}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{11}{12} \mathrm{~L}\)

Explanation:

A Given,
\(\mathrm{m}_{1}=4 \mathrm{~kg}\)
\(\mathrm{~m}_{2}=5 \mathrm{~kg}\)
\(\mathrm{v}_{1}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s} \text { (x direction) }\)
\(\mathrm{v}_{2}=3 \hat{\mathrm{j}} / \mathrm{s} \text { (y direction) }\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{4 \times 5 \hat{\mathrm{i}}+5 \times 3 \hat{\mathrm{j}}}{9}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{20}{9} \hat{\mathrm{i}}+\frac{15}{9} \hat{\mathrm{j}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\left(\frac{20}{9}\right)^{2}+\left(\frac{15}{9}\right)^{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{400+225}{81}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{25}{9} \mathrm{~m} / \mathrm{s}\)

Explanation:

B We know that,
\(\sigma=\frac{\text { mass }}{\text { area }}\)
Mass of the bigger disc-
\(\mathrm{M}_{1}=\sigma \times \pi \times(2 \mathrm{R})^{2}\)
\(\mathrm{M}_{1}=4 \pi \mathrm{R}^{2} \sigma\)
Mass of the smaller disc-
\(\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}^{2}\)

Centre of the mass of cutting disc,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}}{M_{1}+M_{2}}\)
\(X_{c m}=\frac{\left[4 \pi R^{2} \sigma \times 0\right]+\left[\left(-\sigma \times \pi R^{2}\right) \times R\right]}{\left[4 \pi R^{2} \sigma+\left(-\sigma \pi R^{2}\right)\right]}=-\frac{R}{3}\)
\(\alpha R=-\frac{R}{3} \quad\left(\because x_{c m}=\alpha R\right)\)
\(|\alpha|=\left|-\frac{1}{3}\right|\)
\(\alpha=\frac{1}{3}\)

Explanation:

Explanation:

D
Distance from wall \(\left(x_{1}\right)=\frac{L}{2}\)
Distance from wall \(\left(\mathrm{x}_{2}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}=\mathrm{L}\)
Distance from wall \(\left(\mathrm{x}_{3}\right)=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{4}=\frac{5}{4} \mathrm{~L}\)
Now,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}+M_{3} x_{3}}{M_{1}+M_{2}+M_{3}}\)
\(X_{c m}=\frac{M \times \frac{L}{2}+M \times L+M \times \frac{5 L}{4}}{3 M}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{\frac{11}{4} \mathrm{ML}}{3 \mathrm{M}}\)
\(\mathrm{X}_{\mathrm{cm}}=\frac{11}{12} \mathrm{~L}\)

Explanation:

A Given,
\(\mathrm{m}_{1}=4 \mathrm{~kg}\)
\(\mathrm{~m}_{2}=5 \mathrm{~kg}\)
\(\mathrm{v}_{1}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s} \text { (x direction) }\)
\(\mathrm{v}_{2}=3 \hat{\mathrm{j}} / \mathrm{s} \text { (y direction) }\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{4 \times 5 \hat{\mathrm{i}}+5 \times 3 \hat{\mathrm{j}}}{9}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{20}{9} \hat{\mathrm{i}}+\frac{15}{9} \hat{\mathrm{j}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\left(\frac{20}{9}\right)^{2}+\left(\frac{15}{9}\right)^{2}}\)
\(\mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{400+225}{81}}\)
\(\mathrm{v}_{\mathrm{cm}}=\frac{25}{9} \mathrm{~m} / \mathrm{s}\)

Explanation:

B We know that,
\(\sigma=\frac{\text { mass }}{\text { area }}\)
Mass of the bigger disc-
\(\mathrm{M}_{1}=\sigma \times \pi \times(2 \mathrm{R})^{2}\)
\(\mathrm{M}_{1}=4 \pi \mathrm{R}^{2} \sigma\)
Mass of the smaller disc-
\(\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}^{2}\)

Centre of the mass of cutting disc,
\(X_{c m}=\frac{M_{1} x_{1}+M_{2} x_{2}}{M_{1}+M_{2}}\)
\(X_{c m}=\frac{\left[4 \pi R^{2} \sigma \times 0\right]+\left[\left(-\sigma \times \pi R^{2}\right) \times R\right]}{\left[4 \pi R^{2} \sigma+\left(-\sigma \pi R^{2}\right)\right]}=-\frac{R}{3}\)
\(\alpha R=-\frac{R}{3} \quad\left(\because x_{c m}=\alpha R\right)\)
\(|\alpha|=\left|-\frac{1}{3}\right|\)
\(\alpha=\frac{1}{3}\)

Explanation: