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4 months ago

Class 11th Maths

Let E₁ : $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ a > b. Let E₂ be another ellipse such that it touches the end points of major axis of E₁ and the foci of E₂ are the end points of minor axis of E₁. If E₁ and E₂ have same eccentricities, then its value is:

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alok kumar singh

Contributor-Level 10

$E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$e_{1}^{2} = 1 - \frac{b^{2}}{a^{2}}$ -(i)

Let $E_{2} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{B^{2}} = 1$

B > a

$e_{2}^{2} = 1 - \frac{a^{2}}{B^{2}} a n d g i v e n B e_{2} = b$

$e^{2} = \frac{3 - \sqrt[]{5}}{2} = {(\frac{\sqrt[]{5} - 1}{2})}^{2}$

$\Rightarrow e = \frac{\sqrt[]{5} - 1}{2}$

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4 months ago

Class 11th Units and Measurement

Three students S₁, S₂ and S₂ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of Pendulum (cm)	No. of oscillations (n)	Total for n oscillations	Time period (s)
1	64.0	8	128.0	16.0
2	64.0	4	64.0	16.0
3	20.0	4	36.0	9.0

Least count of length = 0.1 cm

least count for time = 0.1 s)

If E₁, E₂ and E₃ are the percentage errors 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no.___________.

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$g = 4 π^{2} \frac{l}{T^{2}}$

$\Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2 Δ T}{T}$

Student 1 will get least error because it has more number of oscillation

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4 months ago

Class 11th System of Particles and Rotational Motion

The centre of a wheel rolling on a plane surface moves with a speed $υ_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt[]{x} υ_{0} .$ Then the value of x is__________.

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Vishal Baghel

Contributor-Level 10

$V_{P} = \sqrt[]{V_{0}^{2} + V_{0}^{2}} = \sqrt[]{2} V_{0}$

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4 months ago

Class 11th Physics System of Particles and Rotational Motion

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter.

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Vishal Baghel

Contributor-Level 10

Using conservation of energy:

$m g h = \frac{1}{2} l_{c m} ω^{2} + \frac{1}{2} m v^{2} = \frac{1}{2} l_{c m} \frac{V^{2}}{r^{2}} + \frac{1}{2} m v^{2}$

$v^{2} = \frac{m g h}{\frac{1}{2} \frac{l_{c m}}{r^{2}} + m v^{2}}$

So, sphere has the greatest velocity of COM and ring has least.

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4 months ago

Class 11th Thermodynamics

What will be the average value of energy for a monatomic gas in thermal equilibrium at temperature T?

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Vishal Baghel

Contributor-Level 10

Total energy of monoatomic gas at equilibrium

$= \frac{3}{2} K_{B} T$

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4 months ago

Class 11th Physics

The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by x(t) = 4 sin $ω t + B c o s ω t w i t h ω = \sqrt[]{\frac{K}{m}} .$ Suppose that at time t = 0, the position of mass is x(0) and velocity $υ (0),$ then its displacement can also be represented as x(t) = C cos $(ω t - ?),$ where C and $?$ are:

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Vishal Baghel

Contributor-Level 10

x (t) = A sin $ω$ t + B cos $ω t$

$\Rightarrow x (t) = \sqrt[]{A^{2} + B^{2}} s i n (ω t + δ) = \sqrt[]{A^{2} + B^{2}} c o s (ω t - (\frac{π}{2} - δ))$

At t = 0:

$x (0) = \sqrt[]{A^{2} + B^{2}} s i n (δ)$

$v (0) = ω \sqrt[]{A^{2} + B^{2}} c o s (δ)$

$\Rightarrow {(x (0))}^{2} + {(\frac{v (0)}{ω})}^{2} = A^{2} + B^{2}$

$t a n ? = t a n (\frac{π}{2} - δ) = c o t δ = \frac{v (0)}{x (0) ω}$

$\Rightarrow ? = t a n^{- 1} (\frac{v (0)}{x (0) ω})$

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4 months ago

Class 11th Maths

The number of solutions of sin⁷ x + cos⁷ x = 1, x $\in [0, 4 π]$ is equal to:

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alok kumar singh

Contributor-Level 10

$s i n^{7} x = c o s^{7} x = 1, x \in [0, 4 π]$

will satisfy for sin x = 1, cos x = 0

$\Rightarrow x = \frac{π}{2} & \frac{5 π}{2} .$

or, cos x = 1, sin x = 0

x = 0, 2π, 4π total 5 solutions

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4 months ago

Class 11th Physics

A bullet of '4g' mass is fired from a gun of mass 4kg. If the bullet moves with the muzzle speed of 50 ms^-¹, the impulse imparted to the gun velocity of recoil of gun are:

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Vishal Baghel

Contributor-Level 10

for bullet

$l = Δ p = m v - 0$

$= \frac{4}{1000} * 50 = \frac{20}{100} = 0.2 k g - m / s$

Using conservation of momentum :

Pi = Pt

$\Rightarrow \frac{4}{1000} * 50 = 4 * v$

v = 0.05 m/s

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4 months ago

Class 11th Maths

Let f : R ® R be defined as f(x) = ${\frac{x^{3}}{{(1 - c o s 2 x)}^{2}} l o g_{e} (\frac{1 + 2 x e^{- 2 x}}{1 - x e^{- x}}) \begin{matrix} , x \neq 0 & , x = 0 \end{matrix} .$ If f is continuous at x = 0, then a is equal to:

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alok kumar singh

Contributor-Level 10

f : R -> R.

$f (x) = [\begin{array}{l} \frac{x^{3}}{{(1 - c o s 2 x)}^{2}} l o g (\frac{1 + 2 x e^{- 2 x}}{{(1 - x e^{- x})}^{2}}) x \neq 0 \\ α x = 0 \end{array}$

As f is continuous at x = 0

$∴ α = \underset{x \to 0}{L i m} f (x)$

$= \underset{x \to 0}{L i m} \frac{1}{2} [e^{- 2 x} + e^{- x}] = \frac{1}{2} * 2 = 1$

$∴ α = 1$

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4 months ago

Class 11th Physics

T₀ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified time period is:

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Vishal Baghel

Contributor-Level 10

$\frac{T}{T'} = \frac{2 π \sqrt[]{\frac{l}{g}}}{2 π \sqrt[]{\frac{l}{\frac{16}{g}}}}$

$\Rightarrow \frac{T}{T'} = \frac{1}{(\frac{1}{4})} = 4$

$\Rightarrow T' = \frac{T}{4}$

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