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1
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be :
A
$${{2MgL} \over {{n^2}}}$$
B
nMgL
C
$${{MgL} \over {2{n^2}}}$$
D
$${{MgL} \over {{n^2}}}$$
2
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y = 0.3 sin(0.157x) cos(200pt). The length of the string is : (All quantities are in SI units.)
A
60 m
B
20 m
C
80 m
D
40 m
3
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Following figure shows two processes A and B for a gas. If $$\Delta $$QA and $$\Delta $$QB are the amount of heat absorbed by the system in two cases, and $$\Delta $$UA and $$\Delta $$UB are changes in internal energies, respectively, then : JEE Main 2019 (Online) 9th April Morning Slot Physics - Heat and Thermodynamics Question 375 English
A
$$\Delta $$QA > $$\Delta $$QB ; $$\Delta $$UA > $$\Delta $$UB
B
$$\Delta $$QA < $$\Delta $$QB ; $$\Delta $$UA < $$\Delta $$UB
C
$$\Delta $$QA > $$\Delta $$QB ; $$\Delta $$UA = $$\Delta $$UB
D
$$\Delta $$QA = $$\Delta $$QB ; $$\Delta $$UA = $$\Delta $$UB
4
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The pressure wave, P = 0.01 sin [1000t – 3x] Nm–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C. On some other day, when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms–1 . Approximate value of T is
A
12°C
B
15°C
C
4°C
D
11°C

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