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PHXII13:NUCLEI

363641 If in a nuclear fission, piece of uranium of mass $0.5 g$ is lost, the energy obtained in $k W h$ is

1

1.25 \times 10^{7}

2

2.25 \times 10^{7}

3

3.25 \times 10^{7}

4

0.25 \times 10^{7}

Explanation:

According to Einstein's mass-energy equivalence relation
$E = m c^{2} = (0.5 \times 10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 4.5 \times 10^{13} J$
$= \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}} k W h (∵ 1 k W h = 3.6 \times 10^{6} J)$
$= 1.25 \times 10^{7} k W h$

PHXII13:NUCLEI

363642 Nuclear reactions obey the law of conservation of

1 Mass and energy

2 Charge

3 Momentum

4 All the above

PHXII13:NUCLEI

363643 If the speed of light were $2 / 3$ of its present value, the energy released in a given atomic explosion will be decreased by a fraction:

1

2 / 3

2

4 / 9

3

3 / 4

4

5 / 9

Explanation:

As we know that, $E = m c^{2}$
$\frac{E_{2}}{E_{1}} = \frac{m {(\frac{2}{3} c)}^{2}}{m c^{2}}$ [mass remain same]
$\frac{E_{2}}{E_{1}} = \frac{4 c^{2}}{q c^{2}} = \frac{4}{9}$ .

PHXII13:NUCLEI

363644 The energy equivalent of $1 g$ of substance is

1

5.6 e V

2

5.6 \times 10^{12} M e V

3

11.2 \times 10^{24} M e V

4

5.6 \times 10^{26} M e V

Explanation:

$E = m c^{2} = 1 \times 10^{- 3} \times {(3 \times 10^{8})}^{2} J$
$= \frac{1 \times 10^{- 3} \times 3 \times 10^{8} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} e V$
$(∵ 1 J = \frac{1}{1.6 \times 10^{- 19}} e V)$
$E = 56.169 \times 10^{31} e V$
$E = 5.6 \times 10^{26} M e V$

JEE - 2024

PHXII13:NUCLEI

363641 If in a nuclear fission, piece of uranium of mass $0.5 g$ is lost, the energy obtained in $k W h$ is

1

1.25 \times 10^{7}

2

2.25 \times 10^{7}

3

3.25 \times 10^{7}

4

0.25 \times 10^{7}

Explanation:

According to Einstein's mass-energy equivalence relation
$E = m c^{2} = (0.5 \times 10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 4.5 \times 10^{13} J$
$= \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}} k W h (∵ 1 k W h = 3.6 \times 10^{6} J)$
$= 1.25 \times 10^{7} k W h$

PHXII13:NUCLEI

363642 Nuclear reactions obey the law of conservation of

1 Mass and energy

2 Charge

3 Momentum

4 All the above

PHXII13:NUCLEI

363643 If the speed of light were $2 / 3$ of its present value, the energy released in a given atomic explosion will be decreased by a fraction:

1

2 / 3

2

4 / 9

3

3 / 4

4

5 / 9

Explanation:

As we know that, $E = m c^{2}$
$\frac{E_{2}}{E_{1}} = \frac{m {(\frac{2}{3} c)}^{2}}{m c^{2}}$ [mass remain same]
$\frac{E_{2}}{E_{1}} = \frac{4 c^{2}}{q c^{2}} = \frac{4}{9}$ .

PHXII13:NUCLEI

363644 The energy equivalent of $1 g$ of substance is

1

5.6 e V

2

5.6 \times 10^{12} M e V

3

11.2 \times 10^{24} M e V

4

5.6 \times 10^{26} M e V

Explanation:

$E = m c^{2} = 1 \times 10^{- 3} \times {(3 \times 10^{8})}^{2} J$
$= \frac{1 \times 10^{- 3} \times 3 \times 10^{8} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} e V$
$(∵ 1 J = \frac{1}{1.6 \times 10^{- 19}} e V)$
$E = 56.169 \times 10^{31} e V$
$E = 5.6 \times 10^{26} M e V$

JEE - 2024

PHXII13:NUCLEI

363641 If in a nuclear fission, piece of uranium of mass $0.5 g$ is lost, the energy obtained in $k W h$ is

1

1.25 \times 10^{7}

2

2.25 \times 10^{7}

3

3.25 \times 10^{7}

4

0.25 \times 10^{7}

Explanation:

According to Einstein's mass-energy equivalence relation
$E = m c^{2} = (0.5 \times 10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 4.5 \times 10^{13} J$
$= \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}} k W h (∵ 1 k W h = 3.6 \times 10^{6} J)$
$= 1.25 \times 10^{7} k W h$

PHXII13:NUCLEI

363642 Nuclear reactions obey the law of conservation of

1 Mass and energy

2 Charge

3 Momentum

4 All the above

PHXII13:NUCLEI

363643 If the speed of light were $2 / 3$ of its present value, the energy released in a given atomic explosion will be decreased by a fraction:

1

2 / 3

2

4 / 9

3

3 / 4

4

5 / 9

Explanation:

As we know that, $E = m c^{2}$
$\frac{E_{2}}{E_{1}} = \frac{m {(\frac{2}{3} c)}^{2}}{m c^{2}}$ [mass remain same]
$\frac{E_{2}}{E_{1}} = \frac{4 c^{2}}{q c^{2}} = \frac{4}{9}$ .

PHXII13:NUCLEI

363644 The energy equivalent of $1 g$ of substance is

1

5.6 e V

2

5.6 \times 10^{12} M e V

3

11.2 \times 10^{24} M e V

4

5.6 \times 10^{26} M e V

Explanation:

$E = m c^{2} = 1 \times 10^{- 3} \times {(3 \times 10^{8})}^{2} J$
$= \frac{1 \times 10^{- 3} \times 3 \times 10^{8} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} e V$
$(∵ 1 J = \frac{1}{1.6 \times 10^{- 19}} e V)$
$E = 56.169 \times 10^{31} e V$
$E = 5.6 \times 10^{26} M e V$

JEE - 2024

PHXII13:NUCLEI

363641 If in a nuclear fission, piece of uranium of mass $0.5 g$ is lost, the energy obtained in $k W h$ is

1

1.25 \times 10^{7}

2

2.25 \times 10^{7}

3

3.25 \times 10^{7}

4

0.25 \times 10^{7}

Explanation:

According to Einstein's mass-energy equivalence relation
$E = m c^{2} = (0.5 \times 10^{- 3} k g) {(3 \times 10^{8} m s^{- 1})}^{2}$
$= 4.5 \times 10^{13} J$
$= \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}} k W h (∵ 1 k W h = 3.6 \times 10^{6} J)$
$= 1.25 \times 10^{7} k W h$

PHXII13:NUCLEI

363642 Nuclear reactions obey the law of conservation of

1 Mass and energy

2 Charge

3 Momentum

4 All the above

PHXII13:NUCLEI

363643 If the speed of light were $2 / 3$ of its present value, the energy released in a given atomic explosion will be decreased by a fraction:

1

2 / 3

2

4 / 9

3

3 / 4

4

5 / 9

Explanation:

As we know that, $E = m c^{2}$
$\frac{E_{2}}{E_{1}} = \frac{m {(\frac{2}{3} c)}^{2}}{m c^{2}}$ [mass remain same]
$\frac{E_{2}}{E_{1}} = \frac{4 c^{2}}{q c^{2}} = \frac{4}{9}$ .

PHXII13:NUCLEI

363644 The energy equivalent of $1 g$ of substance is

1

5.6 e V

2

5.6 \times 10^{12} M e V

3

11.2 \times 10^{24} M e V

4

5.6 \times 10^{26} M e V

Explanation:

$E = m c^{2} = 1 \times 10^{- 3} \times {(3 \times 10^{8})}^{2} J$
$= \frac{1 \times 10^{- 3} \times 3 \times 10^{8} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} e V$
$(∵ 1 J = \frac{1}{1.6 \times 10^{- 19}} e V)$
$E = 56.169 \times 10^{31} e V$
$E = 5.6 \times 10^{26} M e V$

JEE - 2024

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