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PHXII13:NUCLEI

363645 If a proton and antiproton come close to each other and annihilate, energy released will be

1

1.5 \times 10^{- 10} J

2

3 \times 10^{- 10} J

3

4.4 \times 10^{- 10} J

4 None of these

Explanation:

Energy released $= Δ m c^{2} = 2 m_{p} c^{2}$
$= 2 \times 1.67 \times 10^{- 27} \times 9 \times 10^{16}$
$= 3 \times 10^{- 10} J .$
So correct option is (2)

PHXII13:NUCLEI

363646 $1 k g$ of iron (specific heat $120 c a l k g^{- 1} C^{- 1}$ ) is heated by $1000^{\circ} C$ . The increase in its mass is

1 Zero

2

5.6 \times 10^{- 8} k g

3

5.6 \times 10^{- 16} k g

4

5.6 \times 10^{- 12} k g

Explanation:

$Δ m c^{2} = J (M S Δ t)$
$Δ m = \frac{m s (Δ t)}{C^{2}}$
$= \frac{1 \times 120 \times 4.2 \times 1000}{9 \times 10^{16}}$
$= 56 \times 10^{- 13} k g = 5.6 \times 10^{- 12} k g .$

PHXII13:NUCLEI

363647 Mass defect of an atom refers to

1 Inaccurate measurement of mass of neutrons

2 Mass annihilated to produce energy to bind the nucleons

3 Packing fraction

4 Difference in the number of neutrons and protons in the nucleus

PHXII13:NUCLEI

363648 In nuclear fission $0.1 %$ mass is converted into Energy. The energy released in the fission of $1 k g$ mass is

1

2.5 \times 10^{5} k W h

2

2.5 \times 10^{7} k W h

3

2.5 \times 10^{9} k W h

4

2.5 \times 10^{-} 7 k W h

Explanation:

$E = \frac{0.1}{100} m c^{2}$ ; in $k W h = \frac{E}{36 \times 10^{5}}$
$Q (i n k W h) = \frac{0.1}{100} \frac{m c^{2}}{36 \times 10^{5}}$
$= \frac{0.1}{100} \times \frac{1 \times 9 \times 10^{16}}{36 \times 10^{5}}$
$= \frac{0.1}{4} \times 10^{9} = 2.5 \times 10^{7} k W h .$

PHXII13:NUCLEI

363645 If a proton and antiproton come close to each other and annihilate, energy released will be

1

1.5 \times 10^{- 10} J

2

3 \times 10^{- 10} J

3

4.4 \times 10^{- 10} J

4 None of these

Explanation:

Energy released $= Δ m c^{2} = 2 m_{p} c^{2}$
$= 2 \times 1.67 \times 10^{- 27} \times 9 \times 10^{16}$
$= 3 \times 10^{- 10} J .$
So correct option is (2)

PHXII13:NUCLEI

363646 $1 k g$ of iron (specific heat $120 c a l k g^{- 1} C^{- 1}$ ) is heated by $1000^{\circ} C$ . The increase in its mass is

1 Zero

2

5.6 \times 10^{- 8} k g

3

5.6 \times 10^{- 16} k g

4

5.6 \times 10^{- 12} k g

Explanation:

$Δ m c^{2} = J (M S Δ t)$
$Δ m = \frac{m s (Δ t)}{C^{2}}$
$= \frac{1 \times 120 \times 4.2 \times 1000}{9 \times 10^{16}}$
$= 56 \times 10^{- 13} k g = 5.6 \times 10^{- 12} k g .$

PHXII13:NUCLEI

363647 Mass defect of an atom refers to

1 Inaccurate measurement of mass of neutrons

2 Mass annihilated to produce energy to bind the nucleons

3 Packing fraction

4 Difference in the number of neutrons and protons in the nucleus

PHXII13:NUCLEI

363648 In nuclear fission $0.1 %$ mass is converted into Energy. The energy released in the fission of $1 k g$ mass is

1

2.5 \times 10^{5} k W h

2

2.5 \times 10^{7} k W h

3

2.5 \times 10^{9} k W h

4

2.5 \times 10^{-} 7 k W h

Explanation:

$E = \frac{0.1}{100} m c^{2}$ ; in $k W h = \frac{E}{36 \times 10^{5}}$
$Q (i n k W h) = \frac{0.1}{100} \frac{m c^{2}}{36 \times 10^{5}}$
$= \frac{0.1}{100} \times \frac{1 \times 9 \times 10^{16}}{36 \times 10^{5}}$
$= \frac{0.1}{4} \times 10^{9} = 2.5 \times 10^{7} k W h .$

PHXII13:NUCLEI

363645 If a proton and antiproton come close to each other and annihilate, energy released will be

1

1.5 \times 10^{- 10} J

2

3 \times 10^{- 10} J

3

4.4 \times 10^{- 10} J

4 None of these

Explanation:

Energy released $= Δ m c^{2} = 2 m_{p} c^{2}$
$= 2 \times 1.67 \times 10^{- 27} \times 9 \times 10^{16}$
$= 3 \times 10^{- 10} J .$
So correct option is (2)

PHXII13:NUCLEI

363646 $1 k g$ of iron (specific heat $120 c a l k g^{- 1} C^{- 1}$ ) is heated by $1000^{\circ} C$ . The increase in its mass is

1 Zero

2

5.6 \times 10^{- 8} k g

3

5.6 \times 10^{- 16} k g

4

5.6 \times 10^{- 12} k g

Explanation:

$Δ m c^{2} = J (M S Δ t)$
$Δ m = \frac{m s (Δ t)}{C^{2}}$
$= \frac{1 \times 120 \times 4.2 \times 1000}{9 \times 10^{16}}$
$= 56 \times 10^{- 13} k g = 5.6 \times 10^{- 12} k g .$

PHXII13:NUCLEI

363647 Mass defect of an atom refers to

1 Inaccurate measurement of mass of neutrons

2 Mass annihilated to produce energy to bind the nucleons

3 Packing fraction

4 Difference in the number of neutrons and protons in the nucleus

PHXII13:NUCLEI

363648 In nuclear fission $0.1 %$ mass is converted into Energy. The energy released in the fission of $1 k g$ mass is

1

2.5 \times 10^{5} k W h

2

2.5 \times 10^{7} k W h

3

2.5 \times 10^{9} k W h

4

2.5 \times 10^{-} 7 k W h

Explanation:

$E = \frac{0.1}{100} m c^{2}$ ; in $k W h = \frac{E}{36 \times 10^{5}}$
$Q (i n k W h) = \frac{0.1}{100} \frac{m c^{2}}{36 \times 10^{5}}$
$= \frac{0.1}{100} \times \frac{1 \times 9 \times 10^{16}}{36 \times 10^{5}}$
$= \frac{0.1}{4} \times 10^{9} = 2.5 \times 10^{7} k W h .$

PHXII13:NUCLEI

363645 If a proton and antiproton come close to each other and annihilate, energy released will be

1

1.5 \times 10^{- 10} J

2

3 \times 10^{- 10} J

3

4.4 \times 10^{- 10} J

4 None of these

Explanation:

Energy released $= Δ m c^{2} = 2 m_{p} c^{2}$
$= 2 \times 1.67 \times 10^{- 27} \times 9 \times 10^{16}$
$= 3 \times 10^{- 10} J .$
So correct option is (2)

PHXII13:NUCLEI

363646 $1 k g$ of iron (specific heat $120 c a l k g^{- 1} C^{- 1}$ ) is heated by $1000^{\circ} C$ . The increase in its mass is

1 Zero

2

5.6 \times 10^{- 8} k g

3

5.6 \times 10^{- 16} k g

4

5.6 \times 10^{- 12} k g

Explanation:

$Δ m c^{2} = J (M S Δ t)$
$Δ m = \frac{m s (Δ t)}{C^{2}}$
$= \frac{1 \times 120 \times 4.2 \times 1000}{9 \times 10^{16}}$
$= 56 \times 10^{- 13} k g = 5.6 \times 10^{- 12} k g .$

PHXII13:NUCLEI

363647 Mass defect of an atom refers to

1 Inaccurate measurement of mass of neutrons

2 Mass annihilated to produce energy to bind the nucleons

3 Packing fraction

4 Difference in the number of neutrons and protons in the nucleus

PHXII13:NUCLEI

363648 In nuclear fission $0.1 %$ mass is converted into Energy. The energy released in the fission of $1 k g$ mass is

1

2.5 \times 10^{5} k W h

2

2.5 \times 10^{7} k W h

3

2.5 \times 10^{9} k W h

4

2.5 \times 10^{-} 7 k W h

Explanation:

$E = \frac{0.1}{100} m c^{2}$ ; in $k W h = \frac{E}{36 \times 10^{5}}$
$Q (i n k W h) = \frac{0.1}{100} \frac{m c^{2}}{36 \times 10^{5}}$
$= \frac{0.1}{100} \times \frac{1 \times 9 \times 10^{16}}{36 \times 10^{5}}$
$= \frac{0.1}{4} \times 10^{9} = 2.5 \times 10^{7} k W h .$

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