147357
Ratio of density of nuclear matter to density of water is at least $-\left(R_{0}=1.2 \times 10^{-15} \mathrm{~m} \& \mathrm{~m}_{\mathrm{p}}\right.$ $=\mathbf{m}_{\mathbf{n}}=1.67 \times 10^{-27} \mathrm{~kg}$ )
B Given, Radius of nucleus $\left(\mathrm{R}_{0}\right)=1.2 \times 10^{-15} \mathrm{~m}$ Mass of nucleus $\left(\mathrm{M}_{\mathrm{n}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ Volume of nucleus $\left(V_{n}\right)=\frac{4}{3} \pi R^{3}$ $=\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}$ The nuclear density $\rho_{\mathrm{n}}=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho_{\mathrm{n}}=\frac{\mathrm{A} \times 1.67 \times 10^{-27}}{\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27}}{12.56 \times\left(1.2 \times 10^{-15}\right)^{3}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27} \times 10^{45}}{12.56 \times 1.728}$ $=0.2307 \times 10^{18}$ $\rho_{\mathrm{n}}=2.307 \times 10^{17} \mathrm{~kg} . \mathrm{m}^{-3}$
AP EAMCET-06.09.2021
NUCLEAR PHYSICS
147358
The mass number and the volume of a nucleus is $M$ and $V$ respectively. If the mass number is increased to $2 \mathrm{M}$ then the volume is changed to
1 $4 \mathrm{~V}$
2 $\frac{\mathrm{V}}{2}$
3 $2 \mathrm{~V}$
4 $8 \mathrm{~V}$ TS
Explanation:
C Given, Initial mass number $=M$ Initial volume $=\mathrm{V}$ Final mass number $=2 \mathrm{M}$ Density of the nucleus is fixed $\mathrm{M}=\rho \mathrm{V}$ $\mathrm{M} \propto \mathrm{V}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{M}}{2 \mathrm{M}}=\frac{\mathrm{V}}{\mathrm{V}_{2}}$ $\mathrm{~V}_{2}=2 \mathrm{~V}$ $\text { Density }(\rho)=\frac{M}{V}$
EAMCET 05.08.2021
NUCLEAR PHYSICS
147359
The mass number of an iodine nucleus and a polonium nucleus is 125 and 216 respectively. The ratio of radius of iodine nucleus to that of polonium nucleus is
1 $5: 6$
2 $6: 5$
3 $7: 6$
4 $5: 7$ TS
Explanation:
A Given, Mass number of iodine $\left(A_{I}\right)=125$ Mass number of polonium $\left(A_{P}\right)=216$ Then $\frac{R_{I}}{R_{P}}=\left(\frac{A_{I}}{A_{P}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\left(\frac{125}{216}\right)^{1 / 3}$ $=\left\{\frac{(5)^{3}}{(6)^{3}}\right\}^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\frac{5}{6}$ $\mathrm{R}_{\mathrm{I}}: \mathrm{R}_{\mathrm{P}}=5: 6$
147357
Ratio of density of nuclear matter to density of water is at least $-\left(R_{0}=1.2 \times 10^{-15} \mathrm{~m} \& \mathrm{~m}_{\mathrm{p}}\right.$ $=\mathbf{m}_{\mathbf{n}}=1.67 \times 10^{-27} \mathrm{~kg}$ )
B Given, Radius of nucleus $\left(\mathrm{R}_{0}\right)=1.2 \times 10^{-15} \mathrm{~m}$ Mass of nucleus $\left(\mathrm{M}_{\mathrm{n}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ Volume of nucleus $\left(V_{n}\right)=\frac{4}{3} \pi R^{3}$ $=\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}$ The nuclear density $\rho_{\mathrm{n}}=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho_{\mathrm{n}}=\frac{\mathrm{A} \times 1.67 \times 10^{-27}}{\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27}}{12.56 \times\left(1.2 \times 10^{-15}\right)^{3}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27} \times 10^{45}}{12.56 \times 1.728}$ $=0.2307 \times 10^{18}$ $\rho_{\mathrm{n}}=2.307 \times 10^{17} \mathrm{~kg} . \mathrm{m}^{-3}$
AP EAMCET-06.09.2021
NUCLEAR PHYSICS
147358
The mass number and the volume of a nucleus is $M$ and $V$ respectively. If the mass number is increased to $2 \mathrm{M}$ then the volume is changed to
1 $4 \mathrm{~V}$
2 $\frac{\mathrm{V}}{2}$
3 $2 \mathrm{~V}$
4 $8 \mathrm{~V}$ TS
Explanation:
C Given, Initial mass number $=M$ Initial volume $=\mathrm{V}$ Final mass number $=2 \mathrm{M}$ Density of the nucleus is fixed $\mathrm{M}=\rho \mathrm{V}$ $\mathrm{M} \propto \mathrm{V}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{M}}{2 \mathrm{M}}=\frac{\mathrm{V}}{\mathrm{V}_{2}}$ $\mathrm{~V}_{2}=2 \mathrm{~V}$ $\text { Density }(\rho)=\frac{M}{V}$
EAMCET 05.08.2021
NUCLEAR PHYSICS
147359
The mass number of an iodine nucleus and a polonium nucleus is 125 and 216 respectively. The ratio of radius of iodine nucleus to that of polonium nucleus is
1 $5: 6$
2 $6: 5$
3 $7: 6$
4 $5: 7$ TS
Explanation:
A Given, Mass number of iodine $\left(A_{I}\right)=125$ Mass number of polonium $\left(A_{P}\right)=216$ Then $\frac{R_{I}}{R_{P}}=\left(\frac{A_{I}}{A_{P}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\left(\frac{125}{216}\right)^{1 / 3}$ $=\left\{\frac{(5)^{3}}{(6)^{3}}\right\}^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\frac{5}{6}$ $\mathrm{R}_{\mathrm{I}}: \mathrm{R}_{\mathrm{P}}=5: 6$
147357
Ratio of density of nuclear matter to density of water is at least $-\left(R_{0}=1.2 \times 10^{-15} \mathrm{~m} \& \mathrm{~m}_{\mathrm{p}}\right.$ $=\mathbf{m}_{\mathbf{n}}=1.67 \times 10^{-27} \mathrm{~kg}$ )
B Given, Radius of nucleus $\left(\mathrm{R}_{0}\right)=1.2 \times 10^{-15} \mathrm{~m}$ Mass of nucleus $\left(\mathrm{M}_{\mathrm{n}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ Volume of nucleus $\left(V_{n}\right)=\frac{4}{3} \pi R^{3}$ $=\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}$ The nuclear density $\rho_{\mathrm{n}}=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho_{\mathrm{n}}=\frac{\mathrm{A} \times 1.67 \times 10^{-27}}{\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27}}{12.56 \times\left(1.2 \times 10^{-15}\right)^{3}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27} \times 10^{45}}{12.56 \times 1.728}$ $=0.2307 \times 10^{18}$ $\rho_{\mathrm{n}}=2.307 \times 10^{17} \mathrm{~kg} . \mathrm{m}^{-3}$
AP EAMCET-06.09.2021
NUCLEAR PHYSICS
147358
The mass number and the volume of a nucleus is $M$ and $V$ respectively. If the mass number is increased to $2 \mathrm{M}$ then the volume is changed to
1 $4 \mathrm{~V}$
2 $\frac{\mathrm{V}}{2}$
3 $2 \mathrm{~V}$
4 $8 \mathrm{~V}$ TS
Explanation:
C Given, Initial mass number $=M$ Initial volume $=\mathrm{V}$ Final mass number $=2 \mathrm{M}$ Density of the nucleus is fixed $\mathrm{M}=\rho \mathrm{V}$ $\mathrm{M} \propto \mathrm{V}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{M}}{2 \mathrm{M}}=\frac{\mathrm{V}}{\mathrm{V}_{2}}$ $\mathrm{~V}_{2}=2 \mathrm{~V}$ $\text { Density }(\rho)=\frac{M}{V}$
EAMCET 05.08.2021
NUCLEAR PHYSICS
147359
The mass number of an iodine nucleus and a polonium nucleus is 125 and 216 respectively. The ratio of radius of iodine nucleus to that of polonium nucleus is
1 $5: 6$
2 $6: 5$
3 $7: 6$
4 $5: 7$ TS
Explanation:
A Given, Mass number of iodine $\left(A_{I}\right)=125$ Mass number of polonium $\left(A_{P}\right)=216$ Then $\frac{R_{I}}{R_{P}}=\left(\frac{A_{I}}{A_{P}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\left(\frac{125}{216}\right)^{1 / 3}$ $=\left\{\frac{(5)^{3}}{(6)^{3}}\right\}^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\frac{5}{6}$ $\mathrm{R}_{\mathrm{I}}: \mathrm{R}_{\mathrm{P}}=5: 6$
147357
Ratio of density of nuclear matter to density of water is at least $-\left(R_{0}=1.2 \times 10^{-15} \mathrm{~m} \& \mathrm{~m}_{\mathrm{p}}\right.$ $=\mathbf{m}_{\mathbf{n}}=1.67 \times 10^{-27} \mathrm{~kg}$ )
B Given, Radius of nucleus $\left(\mathrm{R}_{0}\right)=1.2 \times 10^{-15} \mathrm{~m}$ Mass of nucleus $\left(\mathrm{M}_{\mathrm{n}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ Volume of nucleus $\left(V_{n}\right)=\frac{4}{3} \pi R^{3}$ $=\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}$ The nuclear density $\rho_{\mathrm{n}}=\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}$ $\rho_{\mathrm{n}}=\frac{\mathrm{A} \times 1.67 \times 10^{-27}}{\frac{4}{3} \pi\left(\mathrm{R}_{0}\right)^{3} \times \mathrm{A}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27}}{12.56 \times\left(1.2 \times 10^{-15}\right)^{3}}$ $\rho_{\mathrm{n}}=\frac{3 \times 1.67 \times 10^{-27} \times 10^{45}}{12.56 \times 1.728}$ $=0.2307 \times 10^{18}$ $\rho_{\mathrm{n}}=2.307 \times 10^{17} \mathrm{~kg} . \mathrm{m}^{-3}$
AP EAMCET-06.09.2021
NUCLEAR PHYSICS
147358
The mass number and the volume of a nucleus is $M$ and $V$ respectively. If the mass number is increased to $2 \mathrm{M}$ then the volume is changed to
1 $4 \mathrm{~V}$
2 $\frac{\mathrm{V}}{2}$
3 $2 \mathrm{~V}$
4 $8 \mathrm{~V}$ TS
Explanation:
C Given, Initial mass number $=M$ Initial volume $=\mathrm{V}$ Final mass number $=2 \mathrm{M}$ Density of the nucleus is fixed $\mathrm{M}=\rho \mathrm{V}$ $\mathrm{M} \propto \mathrm{V}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{M}}{2 \mathrm{M}}=\frac{\mathrm{V}}{\mathrm{V}_{2}}$ $\mathrm{~V}_{2}=2 \mathrm{~V}$ $\text { Density }(\rho)=\frac{M}{V}$
EAMCET 05.08.2021
NUCLEAR PHYSICS
147359
The mass number of an iodine nucleus and a polonium nucleus is 125 and 216 respectively. The ratio of radius of iodine nucleus to that of polonium nucleus is
1 $5: 6$
2 $6: 5$
3 $7: 6$
4 $5: 7$ TS
Explanation:
A Given, Mass number of iodine $\left(A_{I}\right)=125$ Mass number of polonium $\left(A_{P}\right)=216$ Then $\frac{R_{I}}{R_{P}}=\left(\frac{A_{I}}{A_{P}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\left(\frac{125}{216}\right)^{1 / 3}$ $=\left\{\frac{(5)^{3}}{(6)^{3}}\right\}^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{I}}}{\mathrm{R}_{\mathrm{P}}}=\frac{5}{6}$ $\mathrm{R}_{\mathrm{I}}: \mathrm{R}_{\mathrm{P}}=5: 6$