1
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to :
A
$${{3F} \over {\left( {\pi {r^2}YT} \right)}}$$
B
$${{6F} \over {\left( {\pi {r^2}YT} \right)}}$$
C
$${F \over {\left( {3\pi {r^2}YT} \right)}}$$
D
$${9F\left( {\pi {r^2}YT} \right)}$$
2
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) ($$\mu $$0 = 4$$\pi $$ × 10–7 N-A–2) JEE Main 2019 (Online) 12th April Evening Slot Physics - Magnetic Effect of Current Question 133 English
A
1.5 × 10–5 T
B
3.0 × 10–5 T
C
2.0 × 10–5 T
D
2.5 × 10–5 T
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths l1 and l2 where, l1 = nl2 and n is an integer. The ratio k1/k2 of the corresponding force constant, k1 and k2 will be :
A
$${1 \over {{n^2}}}$$
B
$${1 \over n}$$
C
n2
D
n
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass of the planet is $${1 \over 9}$$ th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)
A
$${R \over 9}$$
B
$${R \over 2}$$
C
$${R \over 3}$$
D
$${R \over 4}$$
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