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the distance between two points differing in phase
Question:
The distance between two points differing in phase by 60° on a wave having a wave velocity 360 m/s and frequency 500 Hz is
BCECE - 2017
BCECE
Updated On:
Sep 15, 2024
0.72 m
0.18 m
0.12 m
0.36 m
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The Correct Option is
C
Solution and Explanation
The correct option is (C) : 0.12 m
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Top Questions on distance between two points
Let the line passing through the points
(
−
1
,
2
,
1
)
(-1, 2, 1)
(
−
1
,
2
,
1
)
and parallel to the line
x
−
1
2
=
y
+
1
3
=
z
4
\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}
2
x
−
1
=
3
y
+
1
=
4
z
intersect the line
x
+
2
3
=
y
−
3
2
=
z
−
4
1
\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}
3
x
+
2
=
2
y
−
3
=
1
z
−
4
at the point P. Then the distance of P from the point Q(4, -5, 1) is:
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You measure two quantities as
A
=
1.0
m
±
0.2
m
A = 1.0 \,m \pm 0.2 \,m
A
=
1.0
m
±
0.2
m
,
B
=
2.0
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±
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B = 2.0 \,m \pm 0.2 \,m
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L
1
:
γ
−
=
(
i
+
2
j
+
3
k
)
+
λ
(
i
−
j
+
k
)
L_1: γ^-=(i+2j+3k)+λ(i-j+k)
L
1
:
γ
−
=
(
i
+
2
j
+
3
k
)
+
λ
(
i
−
j
+
k
)
;
L
2
:
γ
−
=
(
4
i
+
5
j
+
6
k
)
+
μ
(
i
+
j
−
k
)
L_2: γ^-=(4i+5j+6k)+μ(i+j-k)
L
2
:
γ
−
=
(
4
i
+
5
j
+
6
k
)
+
μ
(
i
+
j
−
k
)
,
intersect
L
1
L_1
L
1
and
L
2
L_2
L
2
at P and Q respectively. If
(
α
,
β
,
γ
)
(α, β, γ)
(
α
,
β
,
γ
)
is the mid point of the line segment PQ, then
2
(
α
,
β
,
γ
2(α, β, γ
2
(
α
,
β
,
γ
) is equal to
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