1
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 $$\times$$ 103 kg/m3 and its Young’s modulus is 9.27 $$\times$$ 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?
A
7.5 kHz
B
5 kHz
C
2.5 kHz
D
10 kHz
2
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 $$\mu $$m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)
A
100 $$\mu $$m
B
25 $$\mu $$m
C
50 $$\mu $$m
D
75 $$\mu $$m
3
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
An EM wave from air enters a medium. The electric fields are

$$\overrightarrow {{E_1}} $$ = $${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$$ in air and

$$\overrightarrow {{E_2}} $$ = $${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$$ in medium,

where the wave number k and frequency $$\nu $$ refer to their values in air. The medium is non-magnetic. If $${\varepsilon _{{r_1}}}$$ and $${\varepsilon _{{r_2}}}$$ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
A
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$$
B
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$$
C
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$$
D
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$$
4
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
In an a.c. circuit, the instantaneous e.m.f. and current are given by
e = 100 sin 30 t
i = 20 sin $$\left( {30t - {\pi \over 4}} \right)$$
In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively
A
50, 0
B
50, 10
C
$${{1000} \over {\sqrt 2 }},10$$
D
$${{50} \over {\sqrt 2 }}$$
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