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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The Entropy (S) of a black hole can be written as $$S = \beta {k_B}A$$, where kB is the Boltzmann constant and A is the area of the black hole. The $$\beta$$ has dimension of

A
L2
B
ML2L$$-$$1
C
L$$-$$2
D
dimensionless

Explanation

Given, $$S = \beta {k_B}A$$

where, dimensional formula of $$S = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$$

Dimensional formula of $${k_B} = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$$

Dimensional formula of $$A = [{L^2}]$$

Thus, $$[{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}] = \beta [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][{L^2}]$$

$$ \Rightarrow \beta = [{L^{ - 2}}]$$

একটি কৃষ্ণবিবরের (ব্ল্যাক হোলের) এনট্রপি (S) কে $$S = \beta {k_B}A$$ সুত্রদ্বারা প্রকাশ করা যায় (যেখানে kB হল বোলটজমান ধ্রুবক এবং A হল কৃষ্ণবিবরের ক্ষেত্রফল)। সেক্ষেত্রে $$\beta$$ এর মাত্রা হল

A
L2
B
ML2T$$-$$1
C
L$$-$$2
D
মাত্রাহীন
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Given : The percentage error in the measurements of A, B, C and D are respectively, 4%, 2%, 3% and 1%. The relative error in $$Z = {{{A^4}{B^{{1 \over 3}}}} \over {C{D^{{3 \over 2}}}}}$$ is

A
$${{127} \over 2}\% $$
B
$${{127} \over 5}\% $$
C
$${{127} \over 6}\% $$
D
$${{127} \over 7}\% $$

Explanation

The percentage error in Z is given as

$${{\Delta Z} \over Z}\% = 4{{\Delta A} \over A} + {1 \over 3}{{\Delta B} \over B} + {{\Delta C} \over C} + {3 \over 2}{{\Delta D} \over D}$$

$$ = 4 \times 4 + {1 \over 3} \times 2 + 3 + {3 \over 2} \times 1$$

$$ = 16 + {2 \over 3} + 3 + {3 \over 2}$$

$$ = {{127} \over 6}\% $$

A, B, C ও D -এর পরিমাপের শতকরা ত্রুটি যথাক্রমে 4%, 2%, 3% ও 1% হলে $$Z = {{{A^4}{B^{{1 \over 3}}}} \over {C{D^{{3 \over 2}}}}}$$ এর আপেক্ষিক ত্রুটি হবে

A
$${{127} \over 2}\% $$
B
$${{127} \over 5}\% $$
C
$${{127} \over 6}\% $$
D
$${{127} \over 7}\% $$
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
From dimensional analysis, the Rydberg constant can be expressed in terms of electric charge (e), mass (m) and Planck constant (h) as [consider $${1 \over {4\pi {\varepsilon _0}}} \equiv 1$$ unit]
A
$${{{h^2}} \over {m{e^2}}}$$
B
$${{m{e^4}} \over {{h^2}}}$$
C
$${{{m^2}{e^4}} \over {{h^2}}}$$
D
$${{m{e^2}} \over {{h^2}}}$$

Explanation

According to question, from dimensional analysis,

R $$\propto$$ eambhc

$$\Rightarrow$$ R = keambhc

Using dimensional formula of R, e, m and h,

$$[{M^0}{L^{ - 1}}{T^0}] = {[{M^{1/2}}{L^{3/2}}{T^{ - 1}}]^a}{[M]^b}{[M{L^2}{T^{ - 1}}]^c}$$

{Since, $$F = {1 \over {4\pi {\varepsilon _0}}}\,.\,{{{e^2}} \over {{r^2}}} = {{{e^2}} \over {{r^2}}}$$

$$ \Rightarrow {e^2} = F{r^2}$$

$$[e] = {[F{r^2}]^{1/2}} = {[ML{T^{ - 2}}\,.\,{L^2}]^{1/2}} = [{M^{1/2}}{L^{3/2}}{T^{ - 1}}]$$}

$$ \Rightarrow [{M^0}{L^{ - 1}}{T^0}] = [{M^{a/2}}{L^{3a/2}}{T^{ - a}}][{M^b}][{M^c}{L^{2c}}{T^{ - c}}]$$

$$ \Rightarrow [{M^0}{L^{ - 1}}{T^0}] = [{M^{{a \over 2} + b + c}}{L^{{{3a} \over 2} + 2c}}{T^{ - a - c}}]$$

$$\therefore$$ $${a \over 2} + b + c = 0$$ .... (i)

$${{3a} \over 2} + 2c = - 1$$ .... (ii)

$$-$$a $$-$$c = 0 .... (iii)

On solving Eqs. (i), (ii) and (iii), we get

a = 2, b = 1, c = $$-$$2

$$\therefore$$ R $$\propto$$ e2mh$$-$$2

$$ \Rightarrow R \propto {{{e^2}m} \over {{h^2}}}$$
মাত্রার নীতি অনুযায়ী, রিডবার্গ ধ্রুবককে ইলেকট্রনের আধান (e), ভর (m) ও প্লাঙ্কের ধ্রুবকের (h) সমন্বয়ে প্রকাশ করলে তার রাশিমালা হবে ( $${1 \over {4\pi {\varepsilon _0}}} \equiv 1$$ একক ধরে নাও)
A
$${{{h^2}} \over {m{e^2}}}$$
B
$${{m{e^4}} \over {{h^2}}}$$
C
$${{{m^2}{e^4}} \over {{h^2}}}$$
D
$${{m{e^2}} \over {{h^2}}}$$

Explanation

প্রশ্ন অনুসারে, মাত্রিক বিশ্লেষণ থেকে,

R $$\propto$$ eambhc

$$\Rightarrow$$ R = keambhc

R, e, m ও h এর মাত্রিক সূত্র ব্যবহার করে,

$$[{M^0}{L^{ - 1}}{T^0}] = {[{M^{1/2}}{L^{3/2}}{T^{ - 1}}]^a}{[M]^b}{[M{L^2}{T^{ - 1}}]^c}$$

{যেহেতু, $$F = {1 \over {4\pi {\varepsilon _0}}}\,.\,{{{e^2}} \over {{r^2}}} = {{{e^2}} \over {{r^2}}}$$

$$ \Rightarrow {e^2} = F{r^2}$$

$$[e] = {[F{r^2}]^{1/2}} = {[ML{T^{ - 2}}\,.\,{L^2}]^{1/2}} = [{M^{1/2}}{L^{3/2}}{T^{ - 1}}]$$}

$$ \Rightarrow [{M^0}{L^{ - 1}}{T^0}] = [{M^{a/2}}{L^{3a/2}}{T^{ - a}}][{M^b}][{M^c}{L^{2c}}{T^{ - c}}]$$

$$ \Rightarrow [{M^0}{L^{ - 1}}{T^0}] = [{M^{{a \over 2} + b + c}}{L^{{{3a} \over 2} + 2c}}{T^{ - a - c}}]$$

$$\therefore$$ $${a \over 2} + b + c = 0$$ .... (i)

$${{3a} \over 2} + 2c = - 1$$ .... (ii)

$$-$$a $$-$$c = 0 .... (iii)

সমীকরণ (i), (ii) এবং (iii) সমাধান করার সময়, আমরা পাই

a = 2, b = 1, c = $$-$$2

$$\therefore$$ R $$\propto$$ e2mh$$-$$2

$$ \Rightarrow R \propto {{{e^2}m} \over {{h^2}}}$$
4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
The frequency v of the radiation emitted by an atom when an electron jumps from one orbit to another is given by v = k$$\delta $$E, where k is a constant and $$\delta $$E is the change in energy level due to the transition. Then, dimension of k is
A
$${\left[ {M{L^2}{T^{ - 2}}} \right]}$$
B
the same dimension of angular momentum
C
$${\left[ {M{L^2}{T^{ - 1}}} \right]}$$
D
$$\left[ {{M^{ - 1}}{L^{ - 2}}T} \right]$$

Explanation

Given, frequency, v = K$$\delta $$E

where, k is constant and $$\delta $$E is change in energy.

$$ \Rightarrow $$ $$k = {v \over {\delta E}} = {v \over {hv}}$$ .....................[$$ \because $$$$\delta $$E = hv]

= $${1 \over h}$$

$$ \because $$ We know that, energy, E = hv

$$ \Rightarrow $$ $$h = {E \over v} = {{M{L^2}{T^{ - 2}}} \over {{T^{ - 1}}}}$$

h = [ML2T-1]

$$ \therefore $$ Dimension of $$k = {1 \over h} = {1 \over {\left[ {M{L^2}{T^{ - 1}}} \right]}} = \left[ {{M^{ - 1}}{L^{ - 2}}T} \right]$$

কোন পরমাণুর মধ্যে একটি ইলেকট্রন যখন এক কক্ষ থেকে অন্য কক্ষে সংক্রমিত হয় তখন নিঃসৃত বিকিরণের কম্পাঙ্ক যে সমীকরণ মেনে চলে তা হল v = k $$\delta$$E, যেখানে k একটি ধ্রুবক এবং $$\delta$$E হল ওই দুই কক্ষের শক্তির মানের পার্থক্য। তাহলে k-এর মাত্রা হবে --

A
ML2T$$-$$2
B
কৌণিক ভরবেগের মাত্রার সমান
C
ML2T$$-$$1
D
M$$-$$1L$$-$$2T

Explanation

শর্তানুসারে,

v = K$$\delta$$E

$$ \Rightarrow K = {v \over {\delta E}}$$

$$ \Rightarrow [K] = {{[v]} \over {[\delta E]}} = {{[{T^{ - 1}}]} \over {[M{L^2}{T^{ - 2}}]}} = [{M^{ - 1}}{L^{ - 2}}T]$$

$$\Rightarrow$$ Option (d) সঠিক।

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