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physics ncert solutions class 11th

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6 months ago

physics ncert solutions class 11th Class 11th

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of 4 ms^-¹, is__________cm. (take g = 10 ms^-²)

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Vishal Baghel

Contributor-Level 10

.From conservation of energy

mgh = $\frac{1}{2} m v^{2} + \frac{1}{2} l ω^{2}$ $\frac{}{}^{} \frac{}{}^{}$

$\Rightarrow m g h = \frac{1}{2} m v^{2} + \frac{1}{2} \frac{m R^{2}}{2}$

10 h = $\frac{16}{2} + \frac{16}{4}$ $\frac{}{} \frac{}{}$

h = 1.2 m = 120 cm

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6 months ago

physics ncert solutions class 11th Class 11th

The one division of main scale of Vernier calipers reads 1mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale, When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2cm and 6^th Vernier division coincides with a main scale division. The diameter of the bob will be ________* 10^-2 cm.

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Payal Gupta

Contributor-Level 10

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10^-² cm

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6 months ago

physics ncert solutions class 11th Class 11th

An observer is riding on a bicycle and moving towards a hill at 18 kmh^-1. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is 640 Hz and velocity of the sound in air is 320 m/s, the beat frequency between the two sounds heard by observer will be _________ Hz.

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Vishal Baghel

Contributor-Level 10

$v_{S} = 0, v_{O_{b}} = 5 m / s$

$f_{d i r e c t} = (\frac{320 - 5}{320}) 640 = 630 H z$

$f_{b e a t} = (650 - 630) = 20 H z$

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6 months ago

physics ncert solutions class 11th Class 11th

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture of constant volume is $\frac{α^{2}}{4} R J / m o l K;$ then the value of a will be________.

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Vishal Baghel

Contributor-Level 10

$n_{1} = 1 \to m o n o t o m i c$

$n_{2} = 3 \to d i a t o m i c$ $C_{v} = \frac{α^{2} R}{4} J / m o l^{- k}$

As, Cv_(mix) = $\frac{1 * \frac{3}{2} R + 3 * \frac{5}{2} R}{4} = \frac{9 R}{4} = \frac{α^{2} R}{4}$

$∴ α = 3$

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6 months ago

physics ncert solutions class 11th Class 11th

An object is projected in the air with initial velocity u at an angle q. The projectile motion is such that the horizontal range R, in maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be_______degree.

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Vishal Baghel

Contributor-Level 10

$R = \frac{u^{2} s i n (2 * 45 °)}{g} = \frac{u^{2}}{g}$

$\frac{R}{2} = \frac{u^{2}}{2 g} = \frac{u^{2} s i n 20}{g}$

$s i n 2 θ = \frac{1}{2}$

$\Rightarrow 2 θ = 30 ° \Rightarrow θ = 15 °$

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physics ncert solutions class 11th Class 11th

If the projection of $2 \hat{i} + 4 \hat{j} - 2 \hat{k} o n \hat{i} + 2 \hat{j} + α \hat{k}$ is zero. Then, the value of will be________.

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Vishal Baghel

Contributor-Level 10

${\vec{r}}_{1} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}, {\vec{r}}_{2} = \hat{i} + 2 \hat{j} + α \hat{k}$

Projection of ${\vec{r}}_{1}$ $_{}$ on ${\vec{r}}_{2}$ $_{}$ is zero $\Rightarrow {\vec{r}}_{1} . {\vec{r}}_{2}$ $_{}_{}$ = 0

2 + 8 - 2 = 0=> α= 5

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physics ncert solutions class 11th Class 11th

A carnot engine takes 5000 kcal of heat from a reservoir at $727 ° C$ and gives heat to a sink at $127 ° C .$ The work done by the engine is

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Payal Gupta

Contributor-Level 10

T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 * 10³ k cal

$\Rightarrow \frac{600}{1000} = \frac{w}{5000}$

w = 12.6 * 10⁶ J

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6 months ago

physics ncert solutions class 11th Class 11th

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in __________ s.

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Vishal Baghel

Contributor-Level 10

For first ball

s = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow - h = u * 6 - \frac{1}{2} * 10 * 6 * 6$

$\Rightarrow - h = 6 u - 180$

h = 180 - 6u -(1)

for second boy

$h = u t \frac{1}{2} a t^{2}$

$h = u * 1.5 + \frac{1}{2} * 10 * 1.5 * 1.5$

h = 1.5u + 11.25 -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

$u = \frac{168.75}{7.5} = 22.5$

h = 180 - 6u

= 180 - 6 * 22.5

45m

for third ball

$h = u t + \frac{1}{2} a t^{2}$

$h = 0 * t + \frac{1}{2} * 10 t^{2}$

$\Rightarrow 45 = 5 t^{2}$

t² = 9

t = 3 sec

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6 months ago

physics ncert solutions class 11th Class 11th

0.056 kg of Nitrogen is enclose in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is ___________ k cal.

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Vishal Baghel

Contributor-Level 10

$V_{r m s} α \sqrt[]{T} | T_{i} = 127 ° C = 400 k .$

$V_{r m s^{1}} = 2 V_{r m s}, m = 0.056 k g = 56 g$

T_f = 4T_i = 4 * 400 = 1600 k

$Q = n c_{v} Δ T = (\frac{m}{M}) C_{v} Δ T$

= $(\frac{56}{28}) \frac{5}{2} R Δ T$

= $2 * \frac{5}{2} * 2 * 1200$

Q = 12 * 10³ cal

= 12 k cal

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6 months ago

physics ncert solutions class 11th Class 11th

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is :

[Radius of earth R = 6400 km and $\sqrt[]{3} = 1.732]$

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Vishal Baghel

Contributor-Level 10

$g = \frac{9}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow \frac{g}{3} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow {(1 + \frac{h}{R})}^{2} = 3$

1 + $\frac{h}{R} = \sqrt[]{3}$ $\frac{}{}$

$\Rightarrow \frac{h}{R} = \sqrt[]{3} - 1$

$\frac{h}{R} = 1.732 - 1$

$\frac{h}{R} = 0.732$

h = 0.732 * 6400

= 4684.8 km » 4685 km

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