147370
Ratio of charge on positron to mass of positron is approximately
1 $+2 \times 10^{11}$
2 $+5 \times 10^{12}$
3 $-2 \times 10^{11}$
4 $-5 \times 10^{11}$
Explanation:
A Charge of Positron is equal to charge of electron $\mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ $\therefore \quad\frac{\mathrm{q}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.75 \times 10^{+11}$ $= +2 \times 10^{11}$
JIPMER-2019
NUCLEAR PHYSICS
147371
If the radii of ${ }_{30}^{64} \mathrm{Zn}$ and ${ }_{13}^{27} \mathrm{Al}$ nuclei are $\mathrm{R}_{1}$ and $R_{2}$ respectively then $\frac{R_{1}}{R_{2}}=$
1 $\frac{64}{27}$
2 $\frac{4}{3}$
3 $\frac{3}{4}$
4 $\frac{27}{64}$
Explanation:
B Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=27$ We know that volume of the nucleus is directly proportional mass number $\mathrm{V} \propto \mathrm{A}$ $4 / 3 \pi \mathrm{R}^{3} \propto \mathrm{A}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Ratio of radii of $\mathrm{Zn}$ and $\mathrm{Al}$ nuclei are, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{64}{27}\right)^{1 / 3}=\frac{4}{3}$
GUJCET 2019
NUCLEAR PHYSICS
147375
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2: 1$. The ratio of their nuclear sizes will be
1 $2^{\frac{1}{3}}: 1$
2 $1: 3^{\frac{1}{2}}$
3 $1: 2^{\frac{1}{3}}$
4 $3^{\frac{1}{3}}: 1$
Explanation:
C Given that, velocities ratio- $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{2}{1}$ $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ According to conservation of momentum - $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2}\left(-\mathrm{v}_{2}\right)=0$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\therefore \quad \frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ Ratio of there masses, $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\rho \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\rho \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}}=\frac{1}{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}=\frac{1}{2}$ $\therefore \quad\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{1}{2}\right)^{1 / 3}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
TS EAMCET(Medical)-2017
NUCLEAR PHYSICS
147378
Order of magnitude of density of uranium nucleus is $\left(\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}\right)$
1 $10^{20} \mathrm{~kg} / \mathrm{m}^{3}$
2 $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
3 $10^{14} \mathrm{~kg} / \mathrm{m}^{3}$
4 $10^{11} \mathrm{~kg} / \mathrm{m}^{3}$
Explanation:
B Given that, $\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}$ We know that, volume $(\mathrm{V})=\frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{V}=\frac{4}{3} \times 3.14\left[\left(1.2 \times 10^{-15}\right) \mathrm{A}^{1 / 3}\right]^{3}$ $\therefore \quad \rho=\frac{\mathrm{m}}{\mathrm{V}}=\frac{1.67 \times 10^{-27} \mathrm{~kg}}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3} \times \mathrm{A}}$ $\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ Order of density $\approx 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
147370
Ratio of charge on positron to mass of positron is approximately
1 $+2 \times 10^{11}$
2 $+5 \times 10^{12}$
3 $-2 \times 10^{11}$
4 $-5 \times 10^{11}$
Explanation:
A Charge of Positron is equal to charge of electron $\mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ $\therefore \quad\frac{\mathrm{q}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.75 \times 10^{+11}$ $= +2 \times 10^{11}$
JIPMER-2019
NUCLEAR PHYSICS
147371
If the radii of ${ }_{30}^{64} \mathrm{Zn}$ and ${ }_{13}^{27} \mathrm{Al}$ nuclei are $\mathrm{R}_{1}$ and $R_{2}$ respectively then $\frac{R_{1}}{R_{2}}=$
1 $\frac{64}{27}$
2 $\frac{4}{3}$
3 $\frac{3}{4}$
4 $\frac{27}{64}$
Explanation:
B Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=27$ We know that volume of the nucleus is directly proportional mass number $\mathrm{V} \propto \mathrm{A}$ $4 / 3 \pi \mathrm{R}^{3} \propto \mathrm{A}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Ratio of radii of $\mathrm{Zn}$ and $\mathrm{Al}$ nuclei are, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{64}{27}\right)^{1 / 3}=\frac{4}{3}$
GUJCET 2019
NUCLEAR PHYSICS
147375
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2: 1$. The ratio of their nuclear sizes will be
1 $2^{\frac{1}{3}}: 1$
2 $1: 3^{\frac{1}{2}}$
3 $1: 2^{\frac{1}{3}}$
4 $3^{\frac{1}{3}}: 1$
Explanation:
C Given that, velocities ratio- $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{2}{1}$ $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ According to conservation of momentum - $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2}\left(-\mathrm{v}_{2}\right)=0$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\therefore \quad \frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ Ratio of there masses, $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\rho \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\rho \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}}=\frac{1}{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}=\frac{1}{2}$ $\therefore \quad\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{1}{2}\right)^{1 / 3}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
TS EAMCET(Medical)-2017
NUCLEAR PHYSICS
147378
Order of magnitude of density of uranium nucleus is $\left(\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}\right)$
1 $10^{20} \mathrm{~kg} / \mathrm{m}^{3}$
2 $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
3 $10^{14} \mathrm{~kg} / \mathrm{m}^{3}$
4 $10^{11} \mathrm{~kg} / \mathrm{m}^{3}$
Explanation:
B Given that, $\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}$ We know that, volume $(\mathrm{V})=\frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{V}=\frac{4}{3} \times 3.14\left[\left(1.2 \times 10^{-15}\right) \mathrm{A}^{1 / 3}\right]^{3}$ $\therefore \quad \rho=\frac{\mathrm{m}}{\mathrm{V}}=\frac{1.67 \times 10^{-27} \mathrm{~kg}}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3} \times \mathrm{A}}$ $\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ Order of density $\approx 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
147370
Ratio of charge on positron to mass of positron is approximately
1 $+2 \times 10^{11}$
2 $+5 \times 10^{12}$
3 $-2 \times 10^{11}$
4 $-5 \times 10^{11}$
Explanation:
A Charge of Positron is equal to charge of electron $\mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ $\therefore \quad\frac{\mathrm{q}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.75 \times 10^{+11}$ $= +2 \times 10^{11}$
JIPMER-2019
NUCLEAR PHYSICS
147371
If the radii of ${ }_{30}^{64} \mathrm{Zn}$ and ${ }_{13}^{27} \mathrm{Al}$ nuclei are $\mathrm{R}_{1}$ and $R_{2}$ respectively then $\frac{R_{1}}{R_{2}}=$
1 $\frac{64}{27}$
2 $\frac{4}{3}$
3 $\frac{3}{4}$
4 $\frac{27}{64}$
Explanation:
B Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=27$ We know that volume of the nucleus is directly proportional mass number $\mathrm{V} \propto \mathrm{A}$ $4 / 3 \pi \mathrm{R}^{3} \propto \mathrm{A}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Ratio of radii of $\mathrm{Zn}$ and $\mathrm{Al}$ nuclei are, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{64}{27}\right)^{1 / 3}=\frac{4}{3}$
GUJCET 2019
NUCLEAR PHYSICS
147375
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2: 1$. The ratio of their nuclear sizes will be
1 $2^{\frac{1}{3}}: 1$
2 $1: 3^{\frac{1}{2}}$
3 $1: 2^{\frac{1}{3}}$
4 $3^{\frac{1}{3}}: 1$
Explanation:
C Given that, velocities ratio- $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{2}{1}$ $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ According to conservation of momentum - $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2}\left(-\mathrm{v}_{2}\right)=0$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\therefore \quad \frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ Ratio of there masses, $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\rho \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\rho \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}}=\frac{1}{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}=\frac{1}{2}$ $\therefore \quad\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{1}{2}\right)^{1 / 3}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
TS EAMCET(Medical)-2017
NUCLEAR PHYSICS
147378
Order of magnitude of density of uranium nucleus is $\left(\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}\right)$
1 $10^{20} \mathrm{~kg} / \mathrm{m}^{3}$
2 $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
3 $10^{14} \mathrm{~kg} / \mathrm{m}^{3}$
4 $10^{11} \mathrm{~kg} / \mathrm{m}^{3}$
Explanation:
B Given that, $\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}$ We know that, volume $(\mathrm{V})=\frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{V}=\frac{4}{3} \times 3.14\left[\left(1.2 \times 10^{-15}\right) \mathrm{A}^{1 / 3}\right]^{3}$ $\therefore \quad \rho=\frac{\mathrm{m}}{\mathrm{V}}=\frac{1.67 \times 10^{-27} \mathrm{~kg}}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3} \times \mathrm{A}}$ $\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ Order of density $\approx 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
147370
Ratio of charge on positron to mass of positron is approximately
1 $+2 \times 10^{11}$
2 $+5 \times 10^{12}$
3 $-2 \times 10^{11}$
4 $-5 \times 10^{11}$
Explanation:
A Charge of Positron is equal to charge of electron $\mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ $\therefore \quad\frac{\mathrm{q}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.75 \times 10^{+11}$ $= +2 \times 10^{11}$
JIPMER-2019
NUCLEAR PHYSICS
147371
If the radii of ${ }_{30}^{64} \mathrm{Zn}$ and ${ }_{13}^{27} \mathrm{Al}$ nuclei are $\mathrm{R}_{1}$ and $R_{2}$ respectively then $\frac{R_{1}}{R_{2}}=$
1 $\frac{64}{27}$
2 $\frac{4}{3}$
3 $\frac{3}{4}$
4 $\frac{27}{64}$
Explanation:
B Given, $\mathrm{A}_{1}=64, \mathrm{~A}_{2}=27$ We know that volume of the nucleus is directly proportional mass number $\mathrm{V} \propto \mathrm{A}$ $4 / 3 \pi \mathrm{R}^{3} \propto \mathrm{A}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ Ratio of radii of $\mathrm{Zn}$ and $\mathrm{Al}$ nuclei are, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{64}{27}\right)^{1 / 3}=\frac{4}{3}$
GUJCET 2019
NUCLEAR PHYSICS
147375
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2: 1$. The ratio of their nuclear sizes will be
1 $2^{\frac{1}{3}}: 1$
2 $1: 3^{\frac{1}{2}}$
3 $1: 2^{\frac{1}{3}}$
4 $3^{\frac{1}{3}}: 1$
Explanation:
C Given that, velocities ratio- $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{2}{1}$ $\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ According to conservation of momentum - $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2}\left(-\mathrm{v}_{2}\right)=0$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\therefore \quad \frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$ Ratio of there masses, $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\rho \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\rho \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}}=\frac{1}{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}=\frac{1}{2}$ $\therefore \quad\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{1}{2}\right)^{1 / 3}$ $\mathrm{r}_{1}: \mathrm{r}_{2}=1: 2^{1 / 3}$
TS EAMCET(Medical)-2017
NUCLEAR PHYSICS
147378
Order of magnitude of density of uranium nucleus is $\left(\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}\right)$
1 $10^{20} \mathrm{~kg} / \mathrm{m}^{3}$
2 $10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
3 $10^{14} \mathrm{~kg} / \mathrm{m}^{3}$
4 $10^{11} \mathrm{~kg} / \mathrm{m}^{3}$
Explanation:
B Given that, $\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}$ We know that, volume $(\mathrm{V})=\frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{V}=\frac{4}{3} \times 3.14\left[\left(1.2 \times 10^{-15}\right) \mathrm{A}^{1 / 3}\right]^{3}$ $\therefore \quad \rho=\frac{\mathrm{m}}{\mathrm{V}}=\frac{1.67 \times 10^{-27} \mathrm{~kg}}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3} \times \mathrm{A}}$ $\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$ Order of density $\approx 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$
