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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 15
Total: 54
A train S1, moving with a uniform velocity of 108 km/h approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36km/h towards S2 as shown in figure. Both the trains are blowing whistles of same frequency 120Hz. When O is 600m away from S2 and distance between S1 and S2 is 800m, the number of beats heard by O is_____
[ Speed of the sound =330m/s]
Your Answer:
Validate
Solution:
tan
θ
1
=
800
600
=
4
3
θ
1
=
53
∘
θ
2
=
37
∘
⇒ Freq. Heared form train at
S
2
=
f
2
f
2
=
(
V
+
V
0
V
)
f
S
=
(
330
+
10
330
)
×
f
s
=
340
330
f
s
=
1.030
f
s
⇒freq. Heared from train at
S
1
=
f
1
f
1
=
(
V
+
V
0
cos
θ
1
V
−
V
s
cos
θ
2
)
f
s
=
(
330
+
10
×
3
5
330
−
30
×
4
5
)
f
s
f
1
=
(
336
306
)
f
s
=
1.098
f
s
⇒ Beat fieq
=
1.098
f
s
−
1.030
f
s
=
0.068
f
s
=
0.068
×
120
=
8.1
6
≅
8
© examsnet.com
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