1
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

A geostationary satellite is orbiting the Earth at a height of $$4 R$$ above that surface of the Earth. $R$ being the radius of the earth. The, time period of another stellite in coins at a height of $$2 R$$ from the surface of the Earth is.

A
$$2 T_1$$
B
$$2 \sqrt{2} T_1$$
C
$$T_1 / 2$$
D
$$T_1 / 2 \sqrt{2}$$
2
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

Two long parallel wires carry equal current $$i$$ flowing in the same directions are at a distance $$4 d$$ apart. The magnetic field $$B$$ at a point $$P$$ lying on the perpendicular line joining the wires and at a distance $$x$$ from the mid-point is

A
$$\frac{\mu_0 i d}{\pi\left(d^2+x^2\right)}$$
B
$$\frac{\mu_0 i x}{\pi\left(4 d^2-x^2\right)}$$
C
$$\frac{\mu_0 i x}{\left(d^2+x^2\right)}$$
D
$$\frac{\mu_0 i d}{\left(4 d^2+x^2\right)}$$
3
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

0.5 mole of an ideal gas at constant temperature $$27^{\circ} \mathrm{C}$$ kept inside a cylinder of length $$L$$ and cross-section $$A$$ closed by a massless piston. The cylinder is attached with a conducting rod of length L$$_1$$ cross-section area $$(1 / 9) \mathrm{m}^2$$ and thermal conductivity $$k_1$$ whose other end is maintained at $$0^{\circ} \mathrm{C}$$. If piston is moved such that rate of heat flow through the conduction rod is constant then velocity of piston when it is at height $$L / 2$$ from the bottom of cylinder is (neglect any kind of heat loss from system)

VITEEE 2023 Physics - Heat and Thermodynamics Question 10 English

A
$$(\mathrm{k} / R) \mathrm{m} / \mathrm{s}$$
B
$$(\mathrm{k} / 10 R) \mathrm{m} / \mathrm{s}$$
C
$$(\mathrm{k} / 100 R) \mathrm{m} / \mathrm{s}$$
D
$$(\mathrm{k} / 1000 R) \mathrm{m} / \mathrm{s}$$
4
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

In the formula $$X=3 Y Z^2, X$$ and $$Z$$ have dimensions of capacitance and magnetic induction respectively. The dimensions of $$Y$$ is MKS system are.

A
$$\left[M^{-3} L^{-2} T^{-2} A^{-4}\right]$$
B
$$\left[\mathrm{ML}^{-2}\right]$$
C
$$\left[M^{-3} L^{-2} T^8 A^4\right]$$
D
$$\left[M^{-3} L^2 A^4 T^4\right]$$
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