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3 months ago

Class 11th Units and Measurement

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Payal Gupta

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 * 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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3 months ago

Class 11th Physics

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

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Payal Gupta

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow$ $\frac{u_{1}}{u_{2}}$ $=2$

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3 months ago

Class 11th Physics

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

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Payal Gupta

Contributor-Level 10

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

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3 months ago

Class 11th Work, Energy and Power

As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant 2Nm^?1. If both are given velocity in opposite directions, then maximum elongation of the spring is

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Payal Gupta

Contributor-Level 10

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_{r} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{m m}{m + m} = \frac{m}{2}$

By wet

$w_{S P} = Δ K . E .$

$- \frac{1}{2} k x^{2} = 0 - \frac{1}{2} (\frac{m}{2}) {(2 v)}^{2}$

$\frac{2}{2} x^{2} = 0.25 v^{2}$

$x^{2} = 0.25 v^{2}$

$x = \frac{v}{2}$

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3 months ago

Class 11th Physics Gravitation

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth will, be

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Payal Gupta

Contributor-Level 10

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 * \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} * 100 = 1 %$

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3 months ago

Class 11th Physics

A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given p = 3.14)

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Payal Gupta

Contributor-Level 10

$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³= 729r³

$R = (729) \frac{1}{3} (r) (\frac{1}{3})$

R = 9r

$Δ u = T (4 π r^{2}) * 729 - T * 4 π R^{2}$

$= T * 4 π * 8 R^{2}$

$\begin{array}{l} = 75.39 * 1 0^{- 5} J \\ Δ u = 7.5 * 1 0^{- 4} J \end{array}$

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3 months ago

Class 11th Physics Thermodynamics

A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be

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Payal Gupta

Contributor-Level 10

Constant Entropy means Adiabatic process

$P v^{y} = C$

$P_{1} {v_{1}}^{y} = P_{2} {v_{2}}^{y}$

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P2 = 32 P

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3 months ago

Class 11th Physics Thermodynamics

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

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Payal Gupta

Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 * \frac{3}{2} * 8.3 * 40$

$= 3486 J$

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3 months ago

Class 11th Physics Oscillations

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be

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Payal Gupta

Contributor-Level 10

x = A sin $ω t$

$\Rightarrow v = A ω c o s ω t$

$\Rightarrow v = \pm ω \sqrt[]{A^{2} - x^{2}}$

$\frac{v^{2}}{ω^{2}} + x^{2} = A^{2}$

Elliptical

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3 months ago

Class 11th Maths

Let $\frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a},$ $\frac{}{} \frac{}{}$ where a, b, c, are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is:

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Raj Pandey

Contributor-Level 9

$\frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a}$

$= b x d y + c y d y + a d y = a x d x - b y d x + a d x$

$= \frac{c y^{2}}{2} + a y - \frac{a x^{2}}{2} - a x + b x y = k$

$a x^{2} + a y^{2} + 2 a x - 2 a y = k$

$\Rightarrow x^{2} + y^{2} + 2 x - 2 y = λ$

Short distance of (11,6)

= $\sqrt[]{1 2^{2} + 5^{2}} - 5$ $^{}^{}$

= 13 – 5

= 8

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