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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

A golf ball of mass 50 gm placed on a tee, is struck by a golf-club. The speed of the golf ball as it leaves the tee is 100 m/s, the time of contact on the ball is 0.02 s. If the force decreases to zero linearly with time, then the force at the beginning of the contact is

A
100 N
B
200 N
C
250 N
D
500 N

Explanation

Given, initial velocity, $$u = 0$$ m/s

Final velocity, $$v = 100$$ m/s

Mass of ball, $$m = 50 \times {10^{ - 3}}$$ kg

Time of contact $$ = 0.02\,s$$

Impulse, $$F = {{m(v - u)} \over t}$$

$$ = {{50 \times {{10}^{ - 3}}(100 - 0)} \over {0.02}}$$

$$ = {{5 \times 100} \over 2}$$

$$ = 250\,N$$

টি (tee) এর উপর অবস্থিত 50 gm ভরের একটি গলফ বলকে, গলফ ক্লাব দ্বারা আঘাত করা হল। টি ত্যাগ করার সময় গলফ বলের বেগ 100 m/s এবং আঘাতের সংস্পর্শ কাল 0.02s । যদি বলের মান সময়ের সাথে রৈখিক ভাবে কমে শূন্য হয়, তবে শুরুতে প্রযুক্ত বলের মান ছিল

A
100 N
B
200 N
C
250 N
D
500 N
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If a string, suspended from the ceiling is given a downward force F1, its length becomes L1. Its length is L2, if the downward force is F2. What is its actual length?

A
$${{{L_1} + {L_2}} \over 2}$$
B
$$\sqrt {{L_1}{L_2}} $$
C
$${{{F_2}{L_1} + {F_1}{L_2}} \over {{F_2} + {F_1}}}$$
D
$${{{F_2}{L_1} - {F_1}{L_2}} \over {{F_2} - {F_1}}}$$

Explanation

According to the Hooke's law, for deformation of string of length $$l$$,

$${{{F_1}} \over A} \propto {{\Delta {l_1}} \over l}$$ ....... (i)

and in second case,

$${{{F_2}} \over A} \propto {{\Delta {l_2}} \over l}$$ ...... (ii)

From Eqs. (i) and (ii), we have

$${{{F_1}} \over {{F_2}}} = {{\Delta {l_1}} \over {\Delta {l_2}}} = {{{l_1} - l} \over {{l_2} - l}}$$

$$ \Rightarrow ({F_2} - {F_1})l = {F_2}{l_1} - {F_1}{l_2}$$

$$ \Rightarrow l = {{{F_2}{l_1} - {F_1}{l_2}} \over {{F_2} - {F_1}}}$$

সিলিং থেকে ঝোলানো একটি তারকে নিম্নমুখী F1 বল দিয়ে টানলে তার দৈর্ঘ্য হয় L1, F2 বল দিয়ে টানলে তার দৈর্ঘ্য হয় L2 । তাহলে তারটির প্রকৃত দৈর্ঘ্য কত ?

A
$${{{L_1} + {L_2}} \over 2}$$
B
$$\sqrt {{L_1}{L_2}} $$
C
$${{{F_2}{L_1} + {F_1}{L_2}} \over {{F_2} + {F_1}}}$$
D
$${{{F_2}{L_1} - {F_1}{L_2}} \over {{F_2} - {F_1}}}$$
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali

Three blocks are pushed with a force F across a frictionless table as shown in figure above. Let N1 be the contact force between the left two blocks and N2 be the contact force between the right two blocks. Then,
A
F > N1 > N2
B
F > N2 > N1
C
F > N1 = N2
D
F = N1 = N2

Explanation

Given,


Since, the table is frictionless, hence acceleration of the whole system is given as

$$a = {F \over {m + 2m + 3m}} = {F \over {6m}}$$

Free body diagram at 1st block (left most)


From Newton's equation of motion,

$$F - {N_1} = ma = m \times {F \over {6m}}$$ $$\because$$ $$\left( {a = {F \over {6m}}} \right)$$

$$ \Rightarrow {N_1} = {{5F} \over 6}$$

Free body diagram of 3rd block (right most)


From Newton's equation of motion,

$${N_2} = 3ma = 3m \times {F \over {6m}}$$ $$\because$$ $$\left( {a = {F \over {6m}}} \right)$$

$$ \Rightarrow {N_2} = {F \over 2}$$

Hence, F > N1 > N2.

ঘর্ষণহীন একটি টেবিলের উপর রাখা চিত্রে দেখানাে তিনটি ব্লকের উপর F বল প্রয়ােগ করা হল। যদি বামদিকের ব্লকদুটির মধ্যে স্পর্শজনিত বল N1 ও ডানদিকের দুটি ব্লকের মধ্যে স্পর্শজনিত বল N2 হয় তবে
A
F > N1 > N2
B
F > N2 > N1
C
F > N1 = N2
D
F = N1 = N2

Explanation

দেওয়া,


যেহেতু, টেবিলটি ঘর্ষণহীন, তাই পুরো সিস্টেমের ত্বরণ হিসাবে দেওয়া হয়েছে

$$a = {F \over {m + 2m + 3m}} = {F \over {6m}}$$

প্রথম ব্লকে ফ্রি বডি ডায়াগ্রাম (সবচেয়ে বেশি বাঁদিকে)


নিউটনের গতির সমীকরণ থেকে,

$$F - {N_1} = ma = m \times {F \over {6m}}$$ $$\because$$ $$\left( {a = {F \over {6m}}} \right)$$

$$ \Rightarrow {N_1} = {{5F} \over 6}$$

তৃতীয় ব্লকের ফ্রি বডি ডায়াগ্রাম (সবচেয়ে ডানদিকে)


নিউটনের গতির সমীকরণ থেকে,

$${N_2} = 3ma = 3m \times {F \over {6m}}$$ $$\because$$ $$\left( {a = {F \over {6m}}} \right)$$

$$ \Rightarrow {N_2} = {F \over 2}$$

তাই, F > N1 > N2.
4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
A block of mass m rests on a horizontal table with a coefficient of static friction $$\mu $$. What minimum force must be applied on the block to drag it on the table?
A
$${\mu \over {\sqrt {1 + {\mu ^2}} }}$$ mg
B
$${{\mu - 1} \over {\mu + 1}}$$ mg
C
$${\mu \over {\sqrt {1 - {\mu ^2}} }}$$ mg
D
$$\mu $$ mg

Explanation

The block diagram is as shown below,



According to above diagram,

mg = N + Fsin$$\theta $$

$$ \Rightarrow $$ N = mg $$ - $$ Fsin$$\theta $$ ......(i)

and Fcos$$\theta $$ = $$\mu $$N

$$ \Rightarrow $$ Fcos$$\theta $$ = $$\mu $$(mg $$ - $$ Fsin$$\theta $$).............[using Eq. (i)]

For Fmin'

$${d \over {d\theta }}(\cos \theta + \mu \sin \theta ) = 0$$

$$ \Rightarrow - \sin \theta + \mu \cos \theta = 0$$

$$\tan \theta $$ = $$\mu $$

$$\sin \theta = {\mu \over {\sqrt {1 + {\mu ^2}} }}$$ and $$\cos \theta = {1 \over {\sqrt {1 + {\mu ^2}} }}$$

$$ \therefore $$ $${F_{\min }} = {{\mu mg} \over {{1 \over {\sqrt {1 + {\mu ^2}} }} + {{\mu .\mu } \over {\sqrt {1 + {\mu ^2}} }}}}$$

$$ = {{\mu mg} \over {{{1 + {\mu ^2}} \over {\sqrt {1 + {\mu ^2}} }}}} = {{\mu mg} \over {\sqrt {1 + {\mu ^2}} }}$$

$$ \Rightarrow $$ $${F_{\min }} = {{\mu .mg} \over {\sqrt {1 + {\mu ^2}} }}$$

একটি অনুভূমিক টেবিলের উপরে m ভরের একটি বস্তু রাখা আছে। টেবিল ও বস্তুটির মধ্যে স্থির-ঘর্ষণ গুণাঙ্ক $$\mu$$ । বস্তুটিকে টেবিলের উপরে টেনে সরাতে হলে কমপক্ষে কত বল প্রয়োগ করতে হবে ?

A
$${\mu \over {\sqrt {1 + {\mu ^2}} }}$$ mg
B
$${{\mu - 1} \over {\mu + 1}}$$ mg
C
$${\mu \over {\sqrt {1 - {\mu ^2}} }}$$ mg
D
$$\mu$$ mg

Explanation

সাম্যাবস্থায়,

$$F\sin \theta + R = mg$$

$$ \Rightarrow R = (mg - F\sin \theta )$$

এবং $$F\cos \theta = \mu R$$

$$ \Rightarrow F\cos \theta = \mu (mg - F\sin \theta )$$

$$ \Rightarrow F\cos \theta + \mu F\sin \theta = \mu \,mg$$

$$ \Rightarrow F = {{\mu \,mg} \over {(\cos \theta + \mu \,\sin \theta )}}$$

F-এর মান সর্বনিম্ন হলে, $${{dF} \over {d\theta }} = 0$$ হবে।

$${d \over {d\theta }} \left( {{{\mu \,mg} \over {\cos \theta + \mu \sin \theta }}} \right) = 0$$

$$ \Rightarrow \mu mg\left[ {{{ - ( - \sin \theta + \mu \cos \theta )} \over {{{(\cos \theta + \mu \sin \theta )}^2}}}} \right] = 0$$

$$ \Rightarrow \mu \cos \theta - \sin \theta = 0$$

$$ \Rightarrow \mu = \tan \theta $$

$$\therefore$$ $$\cos \theta = {1 \over {\sqrt {1 + {\mu ^2}} }}$$

$$\therefore$$ $${F_{\min }} = {{\mu \,mg} \over {\cos \theta + \mu \sin \theta }}$$

$$ = {{\mu \,mg/\cos \theta } \over {1 + \mu \tan \theta }}$$

$$ = {{\mu \,mg/\sqrt {1 + {\mu ^2}} } \over {1 + {\mu ^2}}}$$

$$ = {{\mu \,mg} \over {\sqrt {1 + {\mu ^2}} }}$$

$$\Rightarrow$$ Option (A) সঠিক।

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