1
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
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The position vector of a particle of mass $$\mathrm{m}$$ moving with a constant velocity $$\vec{v}$$ is given by $$\vec{r}=x(t) \hat{i}+b \hat{j}$$, where $$\mathrm{b}$$ is a constant. At an instant, $$\vec{r}$$ makes an angle $$\theta$$ with the $$x$$-axis as shown in the figure. The variation of the angular momentum of the particle about the origin with $$\theta$$ will be

WB JEE 2024 Physics - Rotational Motion Question 1 English

A
WB JEE 2024 Physics - Rotational Motion Question 1 English Option 1
B
WB JEE 2024 Physics - Rotational Motion Question 1 English Option 2
C
WB JEE 2024 Physics - Rotational Motion Question 1 English Option 3
D
WB JEE 2024 Physics - Rotational Motion Question 1 English Option 4
2
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
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WB JEE 2024 Physics - Center of Mass Question 1 English

The position of the centre of mass of the uniform plate as shown in the figure is

A
$$\left(-\frac{\mathrm{a}}{2},-\frac{\mathrm{b}}{2}\right)$$
B
$$\left(\frac{\mathrm{a}}{8}, \frac{\mathrm{b}}{8}\right)$$
C
$$\left(-\frac{\mathrm{b}}{6},-\frac{\mathrm{a}}{6}\right)$$
D
$$\left(-\frac{a}{6},-\frac{b}{6}\right)$$
3
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
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WB JEE 2024 Physics - Alternating Current Question 2 English

In a series LCR circuit, the rms voltage across the resistor and the capacitor are $$30 \mathrm{~V}$$ and $$90 \mathrm{~V}$$ respectively. If the applied voltage is $$50 \sqrt{2} \sin \omega t$$, then the peak voltage across the inductor is

A
70 V
B
50 V
C
70$$\sqrt2$$ V
D
50$$\sqrt2$$ V
4
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
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WB JEE 2024 Physics - Rotational Motion Question 2 English

A small ball of mass m is suspended from the ceiling of a floor by a string of length $$\mathrm{L}$$. The ball moves along a horizontal circle with constant angular velocity $$\omega$$, as shown in the figure. The torque about the centre (O) of the horizontal circle is

A
$$\mathrm{mgL} \sin \theta$$
B
$$\mathrm{mg} \mathrm{L}$$
C
$$0$$
D
$$\mathrm{mg} \mathrm{L} \cos \theta$$
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