1
JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The velocity-displacement graph describing the motion of bicycle is shown in the figure.

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English
The acceleration-displacement graph of the bicycle's motion is best described by :
A
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 1
B
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 2
C
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 3
D
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 4
2
JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle $$\theta$$ as shown in figure. The coefficient of kinetic friction is $$\mu$$k. then, the block's acceleration 'a' is given by :

(g is acceleration due to gravity)

JEE Main 2021 (Online) 16th March Morning Shift Physics - Laws of Motion Question 83 English
A
$${F \over m}\cos \theta + {\mu _K}\left( {g - {F \over m}\sin \theta } \right)$$
B
$${F \over m}\cos \theta - {\mu _K}\left( {g - {F \over m}\sin \theta } \right)$$
C
$$-$$$${F \over m}\cos \theta - {\mu _K}\left( {g - {F \over m}\sin \theta } \right)$$
D
$${F \over m}\cos \theta - {\mu _K}\left( {g + {F \over m}\sin \theta } \right)$$
3
JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :
A
9.859 $$\times$$ 10$$-$$2 N
B
0.0314 N
C
9.859 $$\times$$ 10$$-$$4 N
D
6.28 $$\times$$ 10$$-$$3 N
4
JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :
A
$$\sqrt 3 T$$
B
$$\sqrt {{2 \over 3}} T$$
C
$${T \over {\sqrt 3 }}$$
D
$$\sqrt {{3 \over 2}} T$$
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