1
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 $$\mathop A\limits^o $$ is used, the minimum separation between two points, to be seen as distinct, will be :
A
0.12 $$\mu $$m
B
0.38 $$\mu $$m
C
0.24 $$\mu $$m
D
0.48 $$\mu $$m
2
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a double slit experiment, when a thin film of thickness t having refractive index $$\mu $$. is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ($$\lambda $$ is the wavelength of the light used) :
A
$${\lambda \over {2\left( {\mu - 1} \right)}}$$
B
$${\lambda \over {\left( {2\mu - 1} \right)}}$$
C
$${{2\lambda } \over {\left( {\mu - 1} \right)}}$$
D
$${\lambda \over {\left( {\mu - 1} \right)}}$$
3
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude $$\theta $$0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his center of mass moves by a distance $$\ell $$($$\ell $$ << L), is close to;
A
mg$$\ell $$(1 + $$\theta $$02)
B
mg$$\ell $$
C
mg$$\ell $$(1 + $${{\theta _0^2} \over 2}$$)
D
mg$$\ell $$(1 - $$\theta $$02)
4
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2 . If it were launched at an angle $$\theta $$0 with speed v0 then (g = 10 ms–2) :
A
$${\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms-1
B
$${\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms-1
C
$${\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms-1
D
$${\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms-1
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