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Class 11th The d and f Block Elements

What is the correct order of the following elements with respect to their density?

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Vishal Baghel

Contributor-Level 10

Density = $\frac{m a s s}{v o l u m e}$

Cu > Co > Fe > Cr > Zn

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Class 11th Work, Energy and Power

A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ……………… m/s. (Take g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

Let v is velocity at the highest position.

$T_{m a x} = 5 T_{m i n} \Rightarrow m g + \frac{m (v^{2} + 4 g l)}{l} = 5 (\frac{m v^{2}}{l} - m g) \Rightarrow 4 . \frac{v^{2}}{l} = 10 g$

$\Rightarrow v = \sqrt[]{\frac{5}{2} g l} = \sqrt[]{\frac{5}{2} * 10 * 1} = 5 m / s$

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Class 11th Physics

A mono-atomic gas of mass 4.0u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is $\frac{x}{3 R} .$ Value of x is…………….

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Vishal Baghel

Contributor-Level 10

$Δ U + Δ K E = 0$

$\Rightarrow \frac{f}{2} n R Δ T = \frac{1}{2} m v^{2} \Rightarrow Δ T = \frac{m}{n} \frac{v^{2}}{f R} = 4 * 1 0^{- 3} * \frac{3 0^{2}}{3 R} = \frac{3.6}{3 R} K .$

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Class 11th Physics Oscillations

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as $U = \frac{α}{r^{10}} - \frac{β}{r^{5}} - 3$

Where, α and β are positive constants. The equilibrium between two atoms will be ${(\frac{2 α}{β})}^{\frac{a}{b}},$ where α =…………….

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Vishal Baghel

Contributor-Level 10

For equilibrium,

$\frac{d U}{d r} = 0 \Rightarrow \frac{- 10 α}{r^{11}} + \frac{5 β}{r^{6}} = 0 \Rightarrow r = {(\frac{2 α}{β})}^{\frac{1}{5}}$

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Class 11th Maths

Let A(1, 4) and B(1, -5) be two points. Let P be a point on the circle ${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ such that ${(P A)}^{2} + {(P B)}^{2}$ have a maximum value, then the points P, A and B lie on:

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alok kumar singh

Contributor-Level 10

$P A^{2} + P B^{2} = c o s^{2} θ + {(s i n θ - 3)}^{2} + c o s^{2} θ + {(s i n θ + 6)}^{2}$

= 2 cos² θ + 2 sin² θ + 6 sin θ + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin q = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

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Class 11th Physics

The root mean square speed of molecules of a given mass of a gas at 27°C and 1 atmosphere pressure is 20 ms^-1. The root mean square speed of molecules of the gas at 127°C and 2 atmosphere pressure is $\frac{x}{\sqrt[]{3}} m s^{- 1} .$ The value of x will be………………….

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Vishal Baghel

Contributor-Level 10

$T = 27 ° C, P = 1 a t m = 1 0^{5} N / m^{2}, V_{r m s} = 200 m / s$

As we know

$v_{r m s} = \sqrt[]{\frac{3 R T}{M}} \Rightarrow \frac{v_{r m s}}{v'_{r m s}} = \frac{\sqrt[]{\frac{3 R T}{M}}}{\sqrt[]{\frac{3 R T'}{M}}}$

$V'_{r m s} = 200 * \frac{2}{\sqrt[]{3}} = \frac{x}{\sqrt[]{3}}$

x = 400

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Class 11th Physics Motion in Straight Line

An engine of a train, moving with uniform acceleration, passes the signal – post with velocity u and the last compartment with velocity $υ .$ The velocity with which middle point of the train passes the signal post is:

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Vishal Baghel

Contributor-Level 10

Assume the length of train be I and its acceleration be a.

$v^{2} = u^{2} + 2 a l \Rightarrow a l = \frac{v^{2} - u^{2}}{2}$

Velocity when middle point crosses the post,

$V_{m} = \sqrt[]{u^{2} + 2 a \frac{l}{2}} . = \sqrt[]{u^{2} + \frac{v^{2} - u^{2}}{2}} = \sqrt[]{\frac{u^{2} + v^{2}}{2}}$

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Class 11th Physics

Two satellites A and B of masses 200kg and 400kg are revolving round the earth at height of 600km and 1600km respectively. If T_A and T_B are the time periods of A and B respectively then the value of T_B – T_A :

[Given : radius of earth = 6400km, mass of earth = 6 * 10²⁴kg]

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{r^{3}}{G M_{e}}}$

$T_{B} - T_{A} = \frac{2 π}{\sqrt[]{G M_{e}}} (r_{B}^{3 / 2} - r_{A}^{3 / 2}) = \frac{2 * 3.14}{\sqrt[]{6.67 * 1 0^{- 11} * 6 * 1 0^{24}}} [{(8 * 1 0^{6})}^{\frac{3}{2}} - {(7 * 1 0^{6})}^{\frac{3}{2}}]$

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Class 11th Physics

Match List – I with List – II :

List – I List – II

(a) h (Planck's constant) &

...more

Match List – I with List – II :

List – I List – II

(a) h (Planck's constant) (i) $[M L T^{- 1}]$

(b) E (kinetic energy) (ii) $[M L^{2} T^{- 1}]$

(c) V (electric potential) (iii) $[M L^{2} T^{- 2}]$

(d) P (linear momentum) (iv) $[M L^{2} I^{- 1} T^{- 3}]$

Choose the correct answer from the options given below:

less

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Vishal Baghel

Contributor-Level 10

[h] = ML²T^-1

[E] = ML²T^-2

[V] = ML²T^-2C^-1

[P] = MLT^-1

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Class 11th Maths

A natural number has prime factorization given by n = $2^{x} 3^{y} 5^{z},$ where y and z are such that y + z = 5 and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z .$ Then the number of odd divisors of n, including 1, is:

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alok kumar singh

Contributor-Level 10

Given n = 2^x. 3^y. 5^z . (i)

$y + z = 5 & \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

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