OK ..... 😊

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Notes are Added ☺. Check out the description BOX... 😍

@ wah yara

@ ya to kar dia ik or kam bhi kar dain na practice mcqs to lagay way sab video mai un ki key bhi kar dain to sir g salute ho jay ga

@@Mz.creation OK ........

As, flux in the coil is = 20 mWb Jab flux ko reverse krei'n ge to 20 mWb ka change aye ga flux me, jo k EMF induce kre ga ...... So, ∆φ = 20 mWb ∆t = 2 ms N = 100 turns Now, emf = N∆φ/∆t emf = (100) (20 mWb)/ (2ms) emf = 1000 V

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Ok ......... I'll add the Link to the Answer Keys in the lecture description Soon In'sha'Allah! & we'll let you know after adding the link. ☺

The decay constant for lead-210 is 0.033915 per year, if its half-life is 20.4 years.

@ shukran habibi

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Sir plz reply

Yess.... Aap theak hai'n ... ☺👍 Slope K.E and frequency graph = K.E/f = hf/f = h (Plank's constant) And the slope of stopping potential & freq. graph = h/e Where, h = Plank's constant & e = charge of electron

@@hafsasiddiqui5977 replied ...... let me know if you get it,,,,, 🙂

Sir please reply

When the battery is disconnected (removed), Q = constant. Now, if we decrease the distance between the plates of the capacitor, the potential difference,'V', also decreases. BUT the ratio V/d = E remains constant. Also, The relation used for the E.F. b/w the plates: E = Q/Aεo Since, Q, A, and εo all are constants. Therefore, E = constant. Hope So, you'll get it. 😊

No ....😊 Reason: 👉 The emission of gamma rays does NOT alter the number of protons or neutrons in the nucleus. So, atomic no. and mass no. remains the same.

sure In'sha'Allah!☺

Yesss ,,,,, You are Right ........... 🥰 Thanks for pointing out the mistake. Jazakallah!☺

Yess ,,,,,,, Frequency bhi greater hoti hia ........ isi wjha se Energy increase ho jati, & in-turn its momentum also increases. Option (A), (B) both are also correct up to some extent. BUT (C) is more accurate,😇

@@Aimers323 ok sir .Thanks.

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Jazakallah! 🥰 In'sha'Allah!❤

@@Aimers323 sir kindlt har chapter kay 50 mcqs aisay mcqs to most expected type hon mean concept hu jo mdcat kay mcqs mai use ho har chapter kay laiye series banaaa dain reconductt say pahlay

@@Mz.creation Before Re-conduct ..... i'll try to upload 2-3 lectures of this type....... 🥰 Moreover, almost 6-7 lectures of PYQs discussion & 2 lectures of Practice Booster Series (100 + 50 MCQs) are already uploaded......... Check out the playlist section 😇

@@Aimers323 IN SH ALLAH you will get success, may you live long

@@Mz.creation In'sha'Allah! May Allah bless you. Ameen!❤

Sir I am asking you Plz confirm coz in the video u have mentioned kai top mai angle 0 hoga

The correct Formulation for Tension at any point is: T = F - mg cosθ Where, θ = angle b/w weight (mg - always vertically downward) & tension (T - along the string). At Top Point: Both T & mg are downward (θ = 0 deg). So, T = F - mg cos0 = F - mg (minimum) OR F = T + mg (maximum) Similarly, at Bottom: Both T & mg are opposite (θ = 180 deg). So, T = F - mg cos180 = F + mg (maximum) OR F = T - mg (minimum)

Jazakallah!🥰