1
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is
A
2m
B
4m
C
1.5m
D
3.5m
2
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $$\ell $$1, and that below the piston is $$\ell $$2, such that $$\ell $$1 > $$\ell $$2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by :
(R is universal gas constant and g is the acceleration due to gravity)
A
$${{nRT} \over g}\left[ {{{{\ell _1} - {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$$
B
$${{RT} \over g}\left[ {{{2{\ell _1} + {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$$
C
$${{nRT} \over g}\left[ {{1 \over {{\ell _2}}} + {1 \over {{\ell _1}}}} \right]$$
D
$${{RT} \over {ng}}\left[ {{{{\ell _1} - 3{\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$$
3
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is – V0. The threshold frequency for photoelectric emission is :
A
2$$v$$
B
$${4 \over 3}v$$
C
$${{3v} \over 2}$$
D
$${{5v} \over 3}$$
4
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is :

(Take $$\varepsilon $$0 = 8.85 $$ \times $$ 10$$-$$12 $${{{C^2}} \over {N - {m^2}}}$$)
A
9.85 $$ \times $$ 10–10 C
B
8.85 $$ \times $$ 10–10 C
C
6.85 $$ \times $$ 10–10 C
D
7.85 × 10–10 C
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