NEET Test Series from KOTA - 10 Papers In MS WORD
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NUCLEAR PHYSICS
147830
A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be
1 1
2 $\frac{3}{4}$
3 $\frac{7}{8}$
4 $\frac{5}{8}$
Explanation:
C Given, Half life of radio isotope $\left(T_{1 / 2}\right)=5$ years Total time to decay $(\mathrm{t})=15$ years We know that, half lives- $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ The fraction of the atoms decayed will be- $1-\frac{1}{8}=\frac{7}{8}$
AMU-2017
NUCLEAR PHYSICS
147831
The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.
1 $\mathrm{X}$ and $\mathrm{Y}$ have the same decay rate initially
2 $\mathrm{X}$ and $\mathrm{Y}$ decay at the same rate always
3 Y will decay at a faster rate than $\mathrm{X}$
4 X will decay at a faster rate than $\mathrm{Y}$
Explanation:
C We know that, Rate of decay $=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ According to question - $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ $\lambda_{\mathrm{X}}=\ln 2 \lambda_{\mathrm{Y}}$ Hence, $\lambda_{y}>\lambda_{x}$. Initially both of them have the same numbers of atoms the $\mathrm{y}$ will decay at a Faster rate than $\mathrm{X}$ because $\mathrm{Y}$ have higher decay constant than X.
JIPMER-2017
NUCLEAR PHYSICS
147832
Radon-222 has a half-life of 3.8 days. If one starts with $0.064 \mathrm{~kg}$ of radon-222, the quantity of radon-222 left after 19 days will be
1 $0.002 \mathrm{~kg}$
2 $0.062 \mathrm{~kg}$
3 $0.032 \mathrm{~kg}$
4 $0.024 \mathrm{~kg}$
Explanation:
A Given that Half-life of radon $-222\left(\mathrm{~T}_{1 / 2}\right)=3.8$ days Initial quantity of radon $\left(\mathrm{N}_{\mathrm{o}}\right)=0.064 \mathrm{~kg}$ Total time taken $(t)=19$ days Quantity of radioactive substance Radon 222 left, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{N}=0.064 \times \frac{1}{32}$ $\mathrm{N}=0.002 \mathrm{~kg}$
WB JEE 2017
NUCLEAR PHYSICS
147833
When ${ }_{92} \mathrm{U}^{235}$ undergoes fission, about $0.1 \%$ of the original mass is converted into energy. The energy released when $1 \mathrm{~kg}$ of ${ }_{92} \mathrm{U}^{235}$ undergoes fission is
1 $9 \times 10^{11} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $9 \times 10^{15} \mathrm{~J}$
4 $9 \times 10^{18} \mathrm{~J}$
Explanation:
B According to question- $\text { Mass }=\frac{0.1}{100} \times 1 \mathrm{~kg}=10^{-3} \mathrm{~kg}$ We know that, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=10^{-3} \times 9 \times 10^{16}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
147830
A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be
1 1
2 $\frac{3}{4}$
3 $\frac{7}{8}$
4 $\frac{5}{8}$
Explanation:
C Given, Half life of radio isotope $\left(T_{1 / 2}\right)=5$ years Total time to decay $(\mathrm{t})=15$ years We know that, half lives- $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ The fraction of the atoms decayed will be- $1-\frac{1}{8}=\frac{7}{8}$
AMU-2017
NUCLEAR PHYSICS
147831
The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.
1 $\mathrm{X}$ and $\mathrm{Y}$ have the same decay rate initially
2 $\mathrm{X}$ and $\mathrm{Y}$ decay at the same rate always
3 Y will decay at a faster rate than $\mathrm{X}$
4 X will decay at a faster rate than $\mathrm{Y}$
Explanation:
C We know that, Rate of decay $=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ According to question - $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ $\lambda_{\mathrm{X}}=\ln 2 \lambda_{\mathrm{Y}}$ Hence, $\lambda_{y}>\lambda_{x}$. Initially both of them have the same numbers of atoms the $\mathrm{y}$ will decay at a Faster rate than $\mathrm{X}$ because $\mathrm{Y}$ have higher decay constant than X.
JIPMER-2017
NUCLEAR PHYSICS
147832
Radon-222 has a half-life of 3.8 days. If one starts with $0.064 \mathrm{~kg}$ of radon-222, the quantity of radon-222 left after 19 days will be
1 $0.002 \mathrm{~kg}$
2 $0.062 \mathrm{~kg}$
3 $0.032 \mathrm{~kg}$
4 $0.024 \mathrm{~kg}$
Explanation:
A Given that Half-life of radon $-222\left(\mathrm{~T}_{1 / 2}\right)=3.8$ days Initial quantity of radon $\left(\mathrm{N}_{\mathrm{o}}\right)=0.064 \mathrm{~kg}$ Total time taken $(t)=19$ days Quantity of radioactive substance Radon 222 left, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{N}=0.064 \times \frac{1}{32}$ $\mathrm{N}=0.002 \mathrm{~kg}$
WB JEE 2017
NUCLEAR PHYSICS
147833
When ${ }_{92} \mathrm{U}^{235}$ undergoes fission, about $0.1 \%$ of the original mass is converted into energy. The energy released when $1 \mathrm{~kg}$ of ${ }_{92} \mathrm{U}^{235}$ undergoes fission is
1 $9 \times 10^{11} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $9 \times 10^{15} \mathrm{~J}$
4 $9 \times 10^{18} \mathrm{~J}$
Explanation:
B According to question- $\text { Mass }=\frac{0.1}{100} \times 1 \mathrm{~kg}=10^{-3} \mathrm{~kg}$ We know that, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=10^{-3} \times 9 \times 10^{16}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
147830
A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be
1 1
2 $\frac{3}{4}$
3 $\frac{7}{8}$
4 $\frac{5}{8}$
Explanation:
C Given, Half life of radio isotope $\left(T_{1 / 2}\right)=5$ years Total time to decay $(\mathrm{t})=15$ years We know that, half lives- $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ The fraction of the atoms decayed will be- $1-\frac{1}{8}=\frac{7}{8}$
AMU-2017
NUCLEAR PHYSICS
147831
The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.
1 $\mathrm{X}$ and $\mathrm{Y}$ have the same decay rate initially
2 $\mathrm{X}$ and $\mathrm{Y}$ decay at the same rate always
3 Y will decay at a faster rate than $\mathrm{X}$
4 X will decay at a faster rate than $\mathrm{Y}$
Explanation:
C We know that, Rate of decay $=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ According to question - $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ $\lambda_{\mathrm{X}}=\ln 2 \lambda_{\mathrm{Y}}$ Hence, $\lambda_{y}>\lambda_{x}$. Initially both of them have the same numbers of atoms the $\mathrm{y}$ will decay at a Faster rate than $\mathrm{X}$ because $\mathrm{Y}$ have higher decay constant than X.
JIPMER-2017
NUCLEAR PHYSICS
147832
Radon-222 has a half-life of 3.8 days. If one starts with $0.064 \mathrm{~kg}$ of radon-222, the quantity of radon-222 left after 19 days will be
1 $0.002 \mathrm{~kg}$
2 $0.062 \mathrm{~kg}$
3 $0.032 \mathrm{~kg}$
4 $0.024 \mathrm{~kg}$
Explanation:
A Given that Half-life of radon $-222\left(\mathrm{~T}_{1 / 2}\right)=3.8$ days Initial quantity of radon $\left(\mathrm{N}_{\mathrm{o}}\right)=0.064 \mathrm{~kg}$ Total time taken $(t)=19$ days Quantity of radioactive substance Radon 222 left, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{N}=0.064 \times \frac{1}{32}$ $\mathrm{N}=0.002 \mathrm{~kg}$
WB JEE 2017
NUCLEAR PHYSICS
147833
When ${ }_{92} \mathrm{U}^{235}$ undergoes fission, about $0.1 \%$ of the original mass is converted into energy. The energy released when $1 \mathrm{~kg}$ of ${ }_{92} \mathrm{U}^{235}$ undergoes fission is
1 $9 \times 10^{11} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $9 \times 10^{15} \mathrm{~J}$
4 $9 \times 10^{18} \mathrm{~J}$
Explanation:
B According to question- $\text { Mass }=\frac{0.1}{100} \times 1 \mathrm{~kg}=10^{-3} \mathrm{~kg}$ We know that, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=10^{-3} \times 9 \times 10^{16}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
147830
A radioisotope has a half-life of 5 years. The fraction of atoms of this material, that would decay in 15 years would be
1 1
2 $\frac{3}{4}$
3 $\frac{7}{8}$
4 $\frac{5}{8}$
Explanation:
C Given, Half life of radio isotope $\left(T_{1 / 2}\right)=5$ years Total time to decay $(\mathrm{t})=15$ years We know that, half lives- $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ The fraction of the atoms decayed will be- $1-\frac{1}{8}=\frac{7}{8}$
AMU-2017
NUCLEAR PHYSICS
147831
The half-life period of a radioactive element $X$ is same as the mean life of another radioactive element Y. Initially, both of them have the same numbers of atoms then.
1 $\mathrm{X}$ and $\mathrm{Y}$ have the same decay rate initially
2 $\mathrm{X}$ and $\mathrm{Y}$ decay at the same rate always
3 Y will decay at a faster rate than $\mathrm{X}$
4 X will decay at a faster rate than $\mathrm{Y}$
Explanation:
C We know that, Rate of decay $=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ Mean life $(\tau)=\frac{1}{\lambda}$ According to question - $\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ $\lambda_{\mathrm{X}}=\ln 2 \lambda_{\mathrm{Y}}$ Hence, $\lambda_{y}>\lambda_{x}$. Initially both of them have the same numbers of atoms the $\mathrm{y}$ will decay at a Faster rate than $\mathrm{X}$ because $\mathrm{Y}$ have higher decay constant than X.
JIPMER-2017
NUCLEAR PHYSICS
147832
Radon-222 has a half-life of 3.8 days. If one starts with $0.064 \mathrm{~kg}$ of radon-222, the quantity of radon-222 left after 19 days will be
1 $0.002 \mathrm{~kg}$
2 $0.062 \mathrm{~kg}$
3 $0.032 \mathrm{~kg}$
4 $0.024 \mathrm{~kg}$
Explanation:
A Given that Half-life of radon $-222\left(\mathrm{~T}_{1 / 2}\right)=3.8$ days Initial quantity of radon $\left(\mathrm{N}_{\mathrm{o}}\right)=0.064 \mathrm{~kg}$ Total time taken $(t)=19$ days Quantity of radioactive substance Radon 222 left, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{N}=0.064 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{N}=0.064 \times \frac{1}{32}$ $\mathrm{N}=0.002 \mathrm{~kg}$
WB JEE 2017
NUCLEAR PHYSICS
147833
When ${ }_{92} \mathrm{U}^{235}$ undergoes fission, about $0.1 \%$ of the original mass is converted into energy. The energy released when $1 \mathrm{~kg}$ of ${ }_{92} \mathrm{U}^{235}$ undergoes fission is
1 $9 \times 10^{11} \mathrm{~J}$
2 $9 \times 10^{13} \mathrm{~J}$
3 $9 \times 10^{15} \mathrm{~J}$
4 $9 \times 10^{18} \mathrm{~J}$
Explanation:
B According to question- $\text { Mass }=\frac{0.1}{100} \times 1 \mathrm{~kg}=10^{-3} \mathrm{~kg}$ We know that, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=10^{-3} \times 9 \times 10^{16}$ $\mathrm{E}=9 \times 10^{13} \mathrm{~J}$
