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Class 11th 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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Class 11th 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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Class 11th Maths

The sum of the series $\sum_{n = 1}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!}$ is equal to:

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alok kumar singh

Contributor-Level 10

$\sum_{n = 1}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r - 1}{2}$

$N o w \sum_{r = (3, 5, 7, . . . . .)}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!} = \sum_{r = (3, 5, 7 . . . . .)}^{\infty} (\frac{\frac{{(r - 1)}^{2}}{4} + 3 r - 3 + 10}{r!}) = \sum_{r = (3, 5, 7, . . . . . .)}^{\infty} (\frac{r^{2} + 10 r + 29}{4 . r!})$

$= \frac{1}{8} {41 e - \frac{19}{e} - 80) = \frac{41}{8} e - \frac{19}{8} e^{- 1} - 10$

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Class 11th 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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Class 11th Physics

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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Class 11th 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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Class 11th Maths

A seven digit number is formed using digit 3,3,4,4,4,5,5. The probability, that number so formed is divisible by 2, is:

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alok kumar singh

Contributor-Level 10

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $= \frac{6!}{2! 2! 2!}$

Total no. of seven digits nos. = $\frac{7!}{2! 3! 2!} * 1$

Hence Req. prob. $\frac{\frac{6!}{2! 2! 2!}}{\frac{7!}{2! 3! 2!}} = \frac{6! 3!}{2! 7!} = \frac{3}{7}$

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Class 11th Units and Measurement

The period of oscillation of a simple pendulum is $T = 2 π \sqrt[]{\frac{L}{g}} .$ Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be:

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{L}{g}} o r T^{2} = 4 π^{2} (\frac{L}{g})$

$\Rightarrow g = 4 π^{2} (\frac{L}{T^{2}})$

$\Rightarrow \frac{Δ g}{g} % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) % = [\frac{1 m m}{1 m} + \frac{2 * 0.01}{1.95}] * 100 % = (0.001 + 0.0102) * 100 = 1.13 %$

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Class 11th Physics

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

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Vishal Baghel

Contributor-Level 10

According to KTG, the gas exerts pressure because its molecule :

Suffer change in momentum when impinge on the walls of container.

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Class 11th Maths

If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x² + y² = 1 is a circle of radius r, then r is equal to:

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alok kumar singh

Contributor-Level 10

$h = \frac{c o s θ + 3}{2}$

$k = \frac{s i n θ + 2}{2}$

-> $c o s θ = 2 h - 3 & s i n θ = 2 k - 2$

-> ${(h - 3 / 2)}^{2} + {(k - 1)}^{2} = \frac{1}{4}$

$∴$ circle of radius r = $\frac{1}{2}$

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