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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference ($$\delta$$) between the two motion is

A
$$\delta = {\pi \over 3}$$
B
$$\delta = {{2\pi } \over 3}$$
C
$$\delta = \pi $$
D
$$\delta = {\pi \over 2}$$

Explanation

According to the question,

$$A = \sqrt {{A^2} + {A^2} + 2AA\cos \delta } $$

$$ \Rightarrow \cos \delta = - {1 \over 2}$$

or, $$\delta = 120^\circ $$

or, $$ = {{2\pi } \over 3}$$

একটি বস্তুকণা একই অভিমুখে দুটি সরল দোলগতির দ্বারা প্রভাবিত হল যাদের কম্পাঙ্ক ও বিস্তার সমমানের। এই অবস্থায় বস্তুকণার চূড়ান্ত বিস্তার যদি দুই প্রভাবক দোলগতির বিস্তারের সহিত একই মানের হয় তাহলে দুই প্রভাবক সরল দোলগতির ভিতর দশা পার্থক্য হবে

A
$$\delta = {\pi \over 3}$$
B
$$\delta = {{2\pi } \over 3}$$
C
$$\delta = \pi $$
D
$$\delta = {\pi \over 2}$$
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
A simple pendulum, consisting of a small ball of mass m attached to a massless string hanging vertically from the ceiling is oscillating with an amplitude such that Tmax = 2Tmin, where Tmax and Tmin are the maximum and minimum tension in the string, respectively. The value of maximum tension Tmax in the string is
A
$${{3mg} \over 2}$$
B
mg
C
$${{3mg} \over 4}$$
D
3 mg

Explanation

The given situation is shown below


$${T_{\min }} = mg\cos \theta $$ ..... (i)

$${T_{\max }} - mg = {{m{v^2}} \over l}$$ ..... (ii)

By law of conservation of energy,

$$mgh = {1 \over 2}m{v^2}$$

$$ \Rightarrow mg(l - l\cos \theta ) = {1 \over 2}m{v^2}$$ [$$\because$$ $$h = l - l\cos \theta $$]

$$2mg(l - l\cos \theta ) = m{v^2}$$

$$ \Rightarrow 2mgl(1 - \cos \theta ) = m{v^2}$$ ..... (iii)

From Eqs. (ii) and (iii), we get

$${T_{\max }} - mg = {{2mgl(1 - \cos \theta )} \over l}$$

$$ \Rightarrow {T_{\max }} - mg = 2mg(1 - \cos \theta )$$

$$ \Rightarrow 2{T_{\min }} - mg = 2mg(1 - \cos \theta )$$ [$$\because$$ $${T_{\max }} = 2{T_{\min }}$$]

$$ \Rightarrow 2mg\cos \theta - mg = 2mg(1 - \cos \theta )$$ [from Eq. (i)]

$$ \Rightarrow 2\cos \theta - 1 = 2(1 - \cos \theta )$$

$$ \Rightarrow 2\cos \theta - 1 = 2 - 2\cos \theta $$

$$ \Rightarrow 4\cos \theta = 3$$

$$ \Rightarrow \cos \theta = {3 \over 4}$$

$$\therefore$$ $${T_{\max }} = 2{T_{\min }} = 2mg\cos \theta = 2mg \times {3 \over 4} = {3 \over 2}mg$$

$$ \Rightarrow {T_{\max }} = {3 \over 2}mg$$
m ভরের একটি ক্ষুদ্র গােলকের সরল দোলক একটি ভরহীন সুতাের সাহায্যে সিলিং থেকে ঝােলানাে আছে।

দোলকটির দোলনকালে Tmax = 2Tmin হয়, যেখানে Tmax ও Tmin হল যথাক্রমে সুতােটির সর্বোচ্চ ও সর্বনিম্ন টান।

সেক্ষেত্রে সর্বোচ্চ টান Tmax -এর মান হবে
A
$${{3mg} \over 2}$$
B
mg
C
$${{3mg} \over 4}$$
D
3 mg

Explanation

প্রদত্ত পরিস্থিতি নীচে দেখানো হয়েছে


$${T_{\min }} = mg\cos \theta $$ ..... (i)

$${T_{\max }} - mg = {{m{v^2}} \over l}$$ ..... (ii)

শক্তি সংরক্ষণ সূত্র দ্বারা,

$$mgh = {1 \over 2}m{v^2}$$

$$ \Rightarrow mg(l - l\cos \theta ) = {1 \over 2}m{v^2}$$ [$$\because$$ $$h = l - l\cos \theta $$]

$$2mg(l - l\cos \theta ) = m{v^2}$$

$$ \Rightarrow 2mgl(1 - \cos \theta ) = m{v^2}$$ ..... (iii)

সমীকরণ (ii) এবং (iii) থেকে, আমরা পাই

$${T_{\max }} - mg = {{2mgl(1 - \cos \theta )} \over l}$$

$$ \Rightarrow {T_{\max }} - mg = 2mg(1 - \cos \theta )$$

$$ \Rightarrow 2{T_{\min }} - mg = 2mg(1 - \cos \theta )$$ [$$\because$$ $${T_{\max }} = 2{T_{\min }}$$]

$$ \Rightarrow 2mg\cos \theta - mg = 2mg(1 - \cos \theta )$$ [সমীকরণ (i) থেকে]

$$ \Rightarrow 2\cos \theta - 1 = 2(1 - \cos \theta )$$

$$ \Rightarrow 2\cos \theta - 1 = 2 - 2\cos \theta $$

$$ \Rightarrow 4\cos \theta = 3$$

$$ \Rightarrow \cos \theta = {3 \over 4}$$

$$\therefore$$ $${T_{\max }} = 2{T_{\min }} = 2mg\cos \theta = 2mg \times {3 \over 4} = {3 \over 2}mg$$

$$ \Rightarrow {T_{\max }} = {3 \over 2}mg$$
3

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
The bob of a swinging second pendulum (one whose time period is 2 s) has a small speed v0 at its lowest point. It height from this lowest point 2.25 s after passing through it, is given by
A
$${{v_0^2} \over {2g}}$$
B
$${{v_0^2} \over {g}}$$
C
$${{v_0^2} \over {4g}}$$
D
$${{9v_0^2} \over {4g}}$$

Explanation

Time period of second pendulum, T = 2s

Displacement equation of simple pendulum,

Given, $$x = A\sin \omega t$$ ...................(i)

t = 2.25 s = 2 + 0.25

= $$2 + {1 \over 4}$$

$$ \therefore $$ t = $$T + {T \over 8}$$

Velocity of second pendulum,

$$v = {{dx} \over {dt}}$$

= $${d \over {dt}}.(A\sin \omega t)$$ .............[from Eq. (i)]

v = A$$\omega $$ cos $$\omega $$ t

Velocity at t = T,

$$v = A\omega \cos \left( {{{2\pi } \over T}.T} \right) = A\omega = {v_0}$$ .............(ii)

Velocity at $$t = T + {T \over {8'}}$$

$${v_1} = A\omega \cos \omega \left( {T + {T \over 8}} \right)$$

= $$A\omega \cos {{2\pi } \over T}\left( {T + {T \over 8}} \right)$$

= $$A\omega \cos 2\pi .{9 \over 8}$$

= $$A\omega \cos \left( {2\pi + {\pi \over 4}} \right)$$

= $$A\omega \cos {\pi \over 4}$$

= $${{A\omega } \over {\sqrt 2 }}$$

$${v_1} = {{{v_0}} \over {\sqrt 2 }}$$ ...........[from Eq. (ii)]

By equation of motion,

$$v_1^2 = v_0^2 - 2gh$$

$$ \Rightarrow $$ $${\left( {{{{v_0}} \over {\sqrt 2 }}} \right)^2} = v_0^2 - 2gh \Rightarrow h = {{v_0^2} \over {4g}}$$

একটি সেকেন্ড পেন্ডুলামের (যার দোলন কাল 2 s পিণ্ডটি তার সর্বনিম্ন অবস্থান, অতি অল্প দ্রুতি v0 সহ অতিক্রম করে। সেক্ষেত্রে সর্বনিম্ন অবস্থান অতিক্রম করার 2.25 s পড়ে পিণ্ডটির উচ্চতা কত হবে ?

A
$${{v_0^2} \over {2g}}$$
B
$${{v_0^2} \over g}$$
C
$${{v_0^2} \over {4g}}$$
D
$${{9v_0^2} \over {4g}}$$

Explanation

পর্যায়কাল (T) = 2 sec

t1 = 2.25 sec

= (2 + 0.25) sec

$$ = T + {T \over 8}$$

= T + t যেখানে, $$t = {T \over 8}$$

সর্বনিম্ন বিন্দুতে বেগ (v0) = $$\omega$$A

t সময়ে বেগ $$(v) = \omega A\cos wt$$

$$ = \omega A\cos {{2\pi } \over T}.{T \over 8}$$

$$ = \omega A\cos {\pi \over 4}$$

$$ = {{\omega A} \over {\sqrt 2 }} = {{{v_0}} \over {\sqrt 2 }}$$

যান্ত্রিক শক্তির সংরক্ষণ থেকে পাই --

$${K_1} + {U_1} = {K_2} + {U_2}$$

$$ \Rightarrow {1 \over 2}m{v_0}^2 + O = {1 \over 2}m{v^2} + mgh$$

$$ \Rightarrow {v_0}^2 - {v^2} = 2gh$$

$$ \Rightarrow {v_0}^2 - {{{v_0}^2} \over 2} = 2gh$$

$$ \Rightarrow 2gh = {{{v_0}} \over 2}$$

$$ \Rightarrow h = {{{v_0}^2} \over {4g}}$$

$$\Rightarrow$$ Option (c) সঠিক।

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