Energy Conservation MCQ Quiz in বাংলা - Objective Question with Answer for Energy Conservation - বিনামূল্যে ডাউনলোড করুন [PDF]

m₁ = m, m₂ = 2m, u₁ = U, u₂ = U/7, V₁ = 0

law of conservation of momentum:

m₁u₁ + m₂u₂ = m₁V₁ + m₂V₂

\(mU + 2m.\frac{U}{7} = m\left( 0 \right) + 2m\left( {{V_2}} \right)\)

\({V_2} = \frac{{9U}}{{14}}\)

Coefficient of restitution:

\(e = \frac{{velocity\;of\;separation}}{{velocity\;of\;approach}} = \frac{{{V_2} - {V_1}}}{{{u_1} - {u_2}}}\)

E_abs,brake = E_initial – E_final

\({v_1} =2~m/s,\;\;{\omega _1} = \frac{{{V_1}}}{R} = \frac{2}{1} =2~rad/s\)

V₂ = 0, ω₂ = 0

h₁ – h₂ = 0.5 - 0 = 0.5 m

I = 150 kg-m², m = 2000 kg

\({E_{initial}} = \frac{1}{2}I\omega _1^2 + \frac{1}{2}mv_1^2 + mg{h_1}\)

\({E_{final}} = \frac{1}{2}I\omega _2^2 + \frac{1}{2}mv_2^2 + mg{h_2}\)

\(Δ E = {E_{initial}} - {E_{final}} = \frac{1}{2}I(\omega _1^2 - \omega _2^2) + \frac{1}{2}m(v_1^2 - v_2^2) + mg({h_1} - {h_2})\)

\(Δ E = \frac{1}{2} \times 150 \times \left[ {{2^2} - {0^2}} \right] + \frac{1}{2} \times 2000\left[ {{2^2} - {0^2}} \right] + 2000 \times 9.81 \times 0.5\)

ΔE = 14110 J

Explanation:

Analyzing the Given Statements about Energy Conversion Systems

A) Entropy generation always occurs in real systems.

In real systems, processes are irreversible, leading to entropy generation. Therefore, this statement is true.

B) Reversible processes produce maximum work.

Reversible processes are ideal and hypothetical; they produce the maximum possible work as there are no losses due to irreversibility. Hence, this statement is true.

C) Carnot cycle is more efficient than Rankine cycle.

The Carnot cycle is an ideal cycle and represents the maximum possible efficiency for a heat engine operating between two temperature limits. The Rankine cycle, used in practical applications like steam turbines, has lower efficiency due to real-world irreversibilities. Thus, this statement is true.

D) Energy conversion efficiency is always above 90% in mechanical systems.

This statement is not true. While some mechanical systems can have high efficiencies, it is not correct to say that they are always above 90%. There are many factors, including friction, material imperfections, and thermal losses, that can reduce efficiency below 90%.

Therefore, the correct answer is option 1: A, B, and C only. 

Concept:

Angular momentum of the system about its center is conserved.

Calculation:

Given:

R = 1m, M = 0.6 kg, m₁ = m₂ = 65g = 0.065 kg

r₁ = 0.4 m, r₂ = 0.6 m, v₁ = v₂ = 25 m/s

Initially, whole system is at rest, therefore

Initial angular momentum (L_i) = 0

After fixing of guns, let system start rotating with angular velocity ‘ω’.

Then,

Final angular momentum (L_f)

L_f = m₁ u₁ r₁ + m₂ u₂ r₂ + I_ω

\({{\rm{L}}_{\rm{f}}}{\rm{\;}} = {m_1}{v_1}\left( {{r_1} + {r_2}} \right) + \frac{1}{2}M{R^2}.\omega \)

\({L_f} = 0.065 \times 25 \times \left( {0.4 + 0.6} \right) + \frac{1}{2} \times 0.6 \times {1^2} \cdot \omega \)

L_f = 1.625 + 0.3 ω

Conservation of angular momentum,

L_i = L_f

0 = 1.625 + 0.3 ω

Explanation:

For unstable equilibrium, the second derivative should be negative. So for slight displacement, the force should take away the body from the equilibrium position

\(\frac{d^2U}{dx^2}<0\)

For stable equilibrium, the second derivative should be positive. So for slight displacement, the restoring force should act on the body

\(\frac{d^2U}{dx^2}>0\)

For Neutral equilibrium, the second derivative should be zero. So for slight displacement, there is no force

\(\frac{d^2U}{dx^2}=0\)

Types of equilibrium:

Explanation:

A Fuel Cell converts chemical energy from a fuel (such as hydrogen) into electrical energy through an electrochemical reaction.

Additional Information

• Dynamo converts mechanical energy into electrical energy, making it incorrect.

• Heat Exchanger is used for thermal energy transfer rather than electrical conversion.

• Solar Cell converts electromagnetic radiation (solar energy) into electrical energy, not chemical.

Explanation:

Statement 1 is correct: According to the given image, energy efficiency (η) is defined as the ratio of work done to energy used:

$$\eta =$$Work done/Energy used​

Thus, it correctly describes efficiency.

Statement 2 is correct: The image states that energy wasted in the process is given by the equation (1 - η)E, where (1 - η) represents the fraction of energy that is lost as heat or inefficiency.

Statement 3 is correct: The image defines efficacy for lighting devices as the ratio of light output (in lumens) to power consumed (in watts). This means efficacy is an efficiency measurement for lighting devices.

Statement 4 is incorrect: If η = 0, it means no useful work is produced and all energy is wasted, which is the case for a completely inefficient system, not an ideal energy device. An ideal device would have η = 1 (100% efficiency), meaning all energy is converted into useful work.

Concept:

Energy stored in the spring = Change in the kinetic energy of gun

½ kx2 = ½ m(v2 – u2)

Calculation:

Given:

m_b = 0.1 kg, m_g = 1 kg, v_b = 100 m/s, recoil of the gun (x) = 0.02 m

Now,

From law of conservation of momentum

m_bu_b + m_gu_g = m_bv_b + m_gv_g

0 + 0 = 0.1 × 100 + 1 × v_g

∴ v_g = - 10 m/s

Now,

Energy stored in the spring = Change in the kinetic energy of gun

½ kx² = ½ m(v² – u²)

½ k (0.02²) = ½ × 1 ((- 10)² - 0)

∴ k = 250000 N/m²

∴ k = 250 kN/m²

Concept:

Use force balance and apply V = Rω at pure rolling

Since there is no external torque, apply angular momentum conservation.

\(\vec T = \frac{{d\vec L}}{{dt}} = 0\)

\(\overrightarrow {{L_i}} = \overrightarrow {{L_f}}\)

Calculation:

Applying force balance

0 – μmg = ma_cm

a_cm = -μg

Applying torque balance about the centre

μmgr = I_cm × α_cm

\(\mu mgr = \frac{{m{r^2}}}{2}{\alpha _{cm}}\)

\({\alpha _{cm}} = \frac{{2\mu g}}{r}\;\left( 1 \right)\)

Considering angular momentum about an axis passing through A and perpendicular to the paper

\(\overrightarrow {{L_A}} = \overrightarrow {{L_{cm}}} + m\left( {\vec r\; \times \vec v} \right)\)

Initially

\({L_{Ai}} = I{\omega _0} + m{v_0}r\)

\({L_{Ai}} = \frac{{m{r^2}}}{2}{\omega _0} + m{v_0}r\)

Final angular momentum (at pure rolling)

\({L_{Af}} = \frac{{m{r^2}}}{2}\omega + mvr,\;\left( {here\;v = r\omega } \right)\)

\({L_{Af}} = \frac{{3m{r^2}}}{2}\omega \)

\(\omega = {\omega _0} + \;{\alpha _{cm}}t\)

\({L_{Ai}} = {L_{Af}}\) (Since there is no external torque applied)

\(\frac{{m{r^2}}}{2}{\omega _0} + m{v_0}r = \frac{3}{2}m{r^2}\left[ {{\omega _0} + \frac{{2\mu g}}{R}t} \right]\)

\(t = \frac{1}{{3\mu g}}\left[ {{V_0} - r\omega } \right]\;\left( 2 \right)\) (option C)

Velocity at pure rolling

V = V₀ + a_cmt (substituting (1) and (2), we get

\(V = V_0 +\frac{{2{V_0}}}{3r} +\frac{{2{}}}{3} r{\omega _0}\) (option (b))

Explanation:

The initial mechanical energy of the roller coaster at point A is completely gravitational potential energy mgH relative to a zero of potential energy at the bottom.

The energy at point B is partly gravitational potential and partly kinetic energy.

\({E_A} = mgH\)

\({E_A} = mg\frac{H}{3} + \frac{1}{2}mv_B^2\)

According to mechanical energy conservation:

\(mgH = mg\frac{H}{3} + \frac{1}{2}mv_B^2\)

\({v_B} = \sqrt {\frac{{2\left( {mgH - mg\frac{H}{3}} \right)}}{m}} = \sqrt {\frac{{4gH}}{3}} \)

Important Point:

If the question was about to find the speed of the roller coaster car at point C

At point C, there is no potential energy, so that the full initial mechanical energy is transformed into kinetic energy.

\({E_A} = mgH = {E_c} = \frac{1}{2}mv_c^2 \Rightarrow {v_c} = \sqrt {2gH} \)