BCECE MAHASANGRAM 2024 || PHYSICS || ATOMS & DUAL NATURE OF VVI MCQ || CLASS NO-14 || BCECE 2024

In this Video I have Discussed about BCECE MAHASANGRAM 2024 || PHYSICS || ELECTROMAGNETIC INDUCTION VVI MCQ || CLASS NO-13 || BCECE 2024

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[ChandaniKumari-pgup]

Question no 19(A) question 20(D) question 21(C) question 22(D) thankyou so much guru ji ❤❤

[Anjalii__kumarii]

Homework question no. (19)-a
m alpha=4m proton
by lembda= h/mv
Vp/Va=4/1 =4:1 ans
(20). - d
lembda= h/√2mkE
on solving lembda= 1.4×10^-10m
= 1.4 A
Q .( 21)- C
lembda is inversely proportional to √m
l1/l2= √2m/√m = √2:1 ans
Q. (22)-C
Kmax= hc/lembda -Work function
on putting the given value we get~
Kmax= 1.4×10^-19 ans ❤️❤️
Thank u sooo much sir 🫶

[ANJALIKUMARI-kwjs]

Formula sheet plzz sir 😊 exam pass aa chuka hai jldi provide kr dijiye taaki ham revise kr le time se

[NishaKumari-wlup]

Sir jo homework data h na uska bhi solution karnba hi digea ga khuki slove nahi hota h

[Hacker_-vm]

Q no. 19 --> A we know 4× mass of proton = mass of alpha particle


Q no. ---> 20 D
We know, K.E = 1/2 mv^2 by this equation we fine mv and put the value of mv in lemda= h/mv

Q no. 21 C
Q no. 22 C

[JyotiKumari-hzzj]

HOME WORK
(19)=4:1, 20- 1.4A°, 21-√2:1, 22- 1.4×10-19

[positivelight]

Homework completely ❤️💖
Sorry for late complete 😅my health is down... 🥹❤️
Ques-19 option(a) is correct❤️
Lemda =h/mv¹
Lemda alpha =h/4mv²
=h/mv¹=h/4mv²
=v¹=4v²
V¹=4v²
=v¹/v²=4:1
So the a is correct option ❤️
Ques--20--option (d) is correct ❤️
given1/2mv²=80×1.6×10-¹⁹
mv=√2×9×¹⁰-³¹×80×1.6×10-¹⁹
=lemda=h/mv

=1.4Å
So the option (d) is correct❤️

Ques 21--option(c)is correct √2:1 is answer✨💫

Ques22--option (c) is correct
—1.6×10-¹⁹=1.4×10-¹⁹
So the option c is correct answer❤️




Thankyou sir for the best session 😇❤️aaj ka class bhot majedar tha❤️😇because chapter thora easy tha😅
Thankyou sir aapke 2class hone ke baad bhi hmara class lene ke liye❤️😇🙏🏻🫶🏻🥰✨
Har har mahadev 🥰✨🥰🫶🏻❤️💖

[RakeshKumar-pseg]

Sir ham aapko se milna chahta hu 14 july ke ❤❤❤❤

[Madhushreya]

19, aa
With solution: lembda, proton
=h/mv1
Lembda alpha =h/4mv2
h/mv1=h/4mv2
V1=4v2
V1/v2=4:1
20, ddd
Given :1/2mv^2 =80×1.6×10-¹⁹
Mv= root under 2×9×10-31×80×1.6×10-19
So, lembda =h/mv
=6.6×1034/root under2×9×10-31×80×1.6×10-19
1.4Å
22, K.Emax= hc/lembda =fie

=3×10-19-1.6×10-19
=1.4×10-19J
21, root 2:1
Thanks😊

[JyotiKumari-hzzj]

Home work
19-A, 20-D, 21-C, 22-D,
Ajj ka lecture bhout accha tha ❤😊❤