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WBCHSE Class 11 Physics Semester 1: Kinematics Chapter MCQ Test with Solutions
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WBCHSE Class 11 Physic...
WBCHSE Class 11 Physic...
Multiple Choice Questions (MCQ)
Class 11 - Semester II
Science Stream
Semester 1
English Medium
WBCHSE Class 11 Physics Semester 1: Kinematics Chapter MCQ Test with Solutions
Interactive chapter test for WBCHSE Class 11 Physics Semester 1. Test your understanding of instantaneous velocity, projectile trajectories, and Newton's laws with real-time feedback and KaTeX formulas.
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প্রশ্ন 1
MCQ
The position of a particle moving along a straight line is given by $x(t) = 3t^2 - 6t + 4$ meters. At what time $t$ will the velocity of the particle become zero?
A
$$t = 1\text{ s}$$
Correct Option
B
$$t = 2\text{ s}$$
C
$$t = 0.5\text{ s}$$
D
$$t = 3\text{ s}$$
সমাধান :
সঠিক বিকল্প: (A)
The velocity function is the first derivative of position with respect to time:
$$v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t + 4) = 6t - 6$$
Setting $v(t) = 0$ gives:
$$6t - 6 = 0 \implies 6t = 6 \implies t = 1\text{ s}$$
Hence, the velocity becomes zero at $t = 1\text{ second}$.
$$v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t + 4) = 6t - 6$$
Setting $v(t) = 0$ gives:
$$6t - 6 = 0 \implies 6t = 6 \implies t = 1\text{ s}$$
Hence, the velocity becomes zero at $t = 1\text{ second}$.
Your Choice: Option
সমাধান :
সঠিক বিকল্প: (A)
The velocity function is the first derivative of position with respect to time:
$$v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t + 4) = 6t - 6$$
Setting $v(t) = 0$ gives:
$$6t - 6 = 0 \implies 6t = 6 \implies t = 1\text{ s}$$
Hence, the velocity becomes zero at $t = 1\text{ second}$.
$$v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t + 4) = 6t - 6$$
Setting $v(t) = 0$ gives:
$$6t - 6 = 0 \implies 6t = 6 \implies t = 1\text{ s}$$
Hence, the velocity becomes zero at $t = 1\text{ second}$.
প্রশ্ন 2
MCQ
A projectile is launched from ground level with speed $u$ at an angle $\theta$ above the horizontal. The maximum vertical height $H_{\max}$ attained by the projectile is given by:
A
$$H_{\max} = \frac{u^2 \sin^2\theta}{2g}$$
Correct Option
B
$$H_{\max} = \frac{u^2 \sin 2\theta}{g}$$
C
$$H_{\max} = \frac{u \sin\theta}{g}$$
D
$$H_{\max} = \frac{2u \sin\theta}{g}$$
সমাধান :
সঠিক বিকল্প: (A)
Using the third equation of kinematics for vertical motion: $v_y^2 = u_y^2 - 2gH$.
At maximum height, vertical velocity $v_y = 0$, and initial vertical velocity $u_y = u\sin\theta$:
$$0 = (u\sin\theta)^2 - 2gH \implies H = \frac{u^2 \sin^2\theta}{2g}$$
Correct option is (A).
At maximum height, vertical velocity $v_y = 0$, and initial vertical velocity $u_y = u\sin\theta$:
$$0 = (u\sin\theta)^2 - 2gH \implies H = \frac{u^2 \sin^2\theta}{2g}$$
Correct option is (A).
Your Choice: Option
সমাধান :
সঠিক বিকল্প: (A)
Using the third equation of kinematics for vertical motion: $v_y^2 = u_y^2 - 2gH$.
At maximum height, vertical velocity $v_y = 0$, and initial vertical velocity $u_y = u\sin\theta$:
$$0 = (u\sin\theta)^2 - 2gH \implies H = \frac{u^2 \sin^2\theta}{2g}$$
Correct option is (A).
At maximum height, vertical velocity $v_y = 0$, and initial vertical velocity $u_y = u\sin\theta$:
$$0 = (u\sin\theta)^2 - 2gH \implies H = \frac{u^2 \sin^2\theta}{2g}$$
Correct option is (A).
প্রশ্ন 3
MCQ
According to Newton's Universal Law of Gravitation $F = G\frac{m_1 m_2}{r^2}$, what is the dimensional formula of the gravitational constant $G$?
A
$$[M^{-1} L^3 T^{-2}]$$
Correct Option
B
$$[M L^2 T^{-2}]$$
C
$$[M^{-1} L^2 T^{-1}]$$
D
$$[M L T^{-2}]$$
সমাধান :
সঠিক বিকল্প: (A)
Rearranging Newton's formula for $G$:
$$G = \frac{F \cdot r^2}{m_1 m_2}$$
Substituting dimensions:
$$[G] = \frac{[M L T^{-2}] \cdot [L^2]}{[M] \cdot [M]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$$
Therefore, the dimension is $[M^{-1} L^3 T^{-2}]$.
$$G = \frac{F \cdot r^2}{m_1 m_2}$$
Substituting dimensions:
$$[G] = \frac{[M L T^{-2}] \cdot [L^2]}{[M] \cdot [M]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$$
Therefore, the dimension is $[M^{-1} L^3 T^{-2}]$.
Your Choice: Option
সমাধান :
সঠিক বিকল্প: (A)
Rearranging Newton's formula for $G$:
$$G = \frac{F \cdot r^2}{m_1 m_2}$$
Substituting dimensions:
$$[G] = \frac{[M L T^{-2}] \cdot [L^2]}{[M] \cdot [M]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$$
Therefore, the dimension is $[M^{-1} L^3 T^{-2}]$.
$$G = \frac{F \cdot r^2}{m_1 m_2}$$
Substituting dimensions:
$$[G] = \frac{[M L T^{-2}] \cdot [L^2]}{[M] \cdot [M]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$$
Therefore, the dimension is $[M^{-1} L^3 T^{-2}]$.
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