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Physics NEET

What is the weightage of Electric Charges and Fields in CBSE Board exam?

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Pallavi Pathak

Contributor-Level 10

Class 12 physics chapter 1 Electric Charges and Fields is an important chapter for the CBSE Board exam. In unit 1 of the CBSE Board exam of class 12 Physics, there are questions from three chapters including - Electric Charges and Fields, Electrostatics, and Electrostatic Potential and Capacitance. This unit carries a total 16 marks.

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3 months ago

Physics Physics Oscillations

A student is performing the experiment of resonance column. The diameter of the column tube is 6cm. THE Frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336m/s. The zero of the meter scale coincides with tip end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:

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Vishal Baghel

Contributor-Level 10

For first resonance,

$l + 0.3 d = \frac{v}{4 f}$

$\Rightarrow l + 0.3 * 6 = \frac{336 * 100}{4 * 504} \Rightarrow l = 14.8 c m$

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3 months ago

Physics Physics Moving Charges and Magnetism

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05m 0.2m from the centre are in the ratio 8 : 1. The radius of coil is-------------.

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Vishal Baghel

Contributor-Level 10

Magnetic field on the axis of a circular coil at distance x from centre, B = $\frac{μ_{0} N i r^{2}}{2 {(r^{2} + x^{2})}^{\frac{3}{2}}}$

$\frac{{(r^{2} + {(0.2)}^{2})}^{\frac{3}{2}}}{{(r^{2} + {(0.05)}^{2})}^{\frac{3}{2}}} = 8$ $\frac{r^{2} + {(0.2)}^{2}}{r^{2} + {(0.05)}^{2}} = 4 \Rightarrow r^{2} = 0.01 \Rightarrow r = 0.1 m .$

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Physics Class 12th

Two radioactive substances X and Y originally have N₁ and N₂ nuclei respectively. Half life of X is half of the life of Y. After three half lives of Y, number of nuclei of both are equal.

The ratio $\frac{N_{1}}{N_{2}}$ will be equal to:

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Vishal Baghel

Contributor-Level 10

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_{1}}{2^{6}} = \frac{N_{2}}{2^{3}} \Rightarrow \frac{N_{1}}{N_{2}} = 8$

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3 months ago

Physics Oscillations

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

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3 months ago

Physics Thermodynamics

A diatomic gas, having C_P $= \frac{7}{2} R a n d C_{V} = \frac{5}{2} R,$ is heated at constant pressure. The ratio dU : dQ : dW :

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Vishal Baghel

Contributor-Level 10

At constant pressure,

$d U = n C_{V} d T = n . \frac{5}{2} R d T$

$d Q = n C_{P} d T = n . \frac{7}{2} R d T$

dW = nRdT

dU : dQ : dW = 5 : 7 : 2

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3 months ago

Physics Class 12th

A 5 V battery is connected across the points X and Y. Assume D₁ and D₂ to the normal silicon diodes. Find the current supplies by the battery if the +ve terminal of the battery is connected to point X.

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Vishal Baghel

Contributor-Level 10

D₁ is in forward bias and D₂ is in reverse bias.

Current, I $= \frac{5 - 0.7}{10} = 0.43 A$

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Physics Class 12th

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Physics Class 11th

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.

Reason R : On heating, the length of the rod increases.

In the light of the above statements, choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Thermal stress is developed on heating when expansion of rod is hindered.

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3 months ago

Physics Units and Measurement

The pitch of the screw gauge is 1mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72^nd division on circular scale coincides with the reference line, The radius of the wire is:

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Vishal Baghel

Contributor-Level 10

$L C = \frac{1}{100} m m = 0.01 m m$

Zero error = +0.08 mm.

Diameter = 1 + 72 * 0.01 – 0.08 = 1.64 mm

Radius = 0.82 mm

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