QBManager

WAVES

172814 Wire having tension $225 N$ produces six beats per second when it is tuned with a fork. When tension changes to $256 N$ , it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

1

186 Hz

2

225 Hz

3

256 Hz

4

280 Hz

Explanation:

A For wire, vibrating under tension, the fundamental frequency is given by
$f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
Where, $T$ is tension $& μ$ is mass per unit length of the string.
It is given that, when it is sounded with a tuning fork of frequency $x, 6$ beats per seconds were heard
$∴ (x - f_{1}) = 6$
$f_{1} = x - 6$
$∴ f_{1} = \frac{1}{2 l} \sqrt{\frac{225}{μ}}; f_{2} = \frac{1}{2 l} \sqrt{\frac{256}{μ}}$
$\frac{f_{1}}{f_{2}} = \frac{15}{16}$
$f_{2} = \frac{16}{15} \times f_{1} = \frac{16}{15} (x - 6)$
$But f_{2} - x = 6$
$f_{2} = (x + 6)$
$∴ (x + 6) = \frac{16}{15} (x - 6)$
$15 x + 90 = 16 x - 96$
$x = 186 Hz$

MHT-CET 2016

WAVES

172815 A source of unknown frequency gives 4 beats $s^{- 1}$ when sounded with a source of known frequency $250 Hz$ . The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 Hz$ . The unknown frequency is

1

254 Hz

2

246 Hz

3

240 Hz

4

260 Hz

Explanation:

A $f_{known} = 250 Hz$
$n = 4 beats / sec$
$f_{unknown} = 250 \pm 4$
$= 246 Hz or 254 Hz$
Now, the second harmonic of the source of unknown frequency, $2 f = 513 \pm 5$
$= 513 \pm 5$
$= 518 Hz or 508 Hz$
$∴ I^{st}$ harmonic $= \frac{518}{2}$ or $\frac{508}{2}$
$= 259 or 254$
Hence, the unknown frequency $= 254 Hz$

MHT-CET 2013

WAVES

172817 A note has a frequency $128 Hz$ . The frequency of a note two octaves higher than it is

1

256 Hz

2

64 Hz

3

32 Hz

4

512 Hz

Explanation:

D We know
$n = \log_{2} (\frac{f_{2}}{f_{1}})$
$2 = \log_{2} (\frac{f_{2}}{f_{1}})$
$2^{2} = \frac{f_{2}}{f_{1}}$
$f_{2} = 4 \times f_{1}$
$f_{2} = 4 \times 128$
$f_{2} = 512 Hz = 2$

MHT-CET 2006

WAVES

172818 Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

B We know that,
$f = 600 Hz$
$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
If increase in tension of one wires then beats per second is -
$f + 6 = \frac{1}{2 l} \sqrt{\frac{T^{'}}{m}}$
Dividing the equation (i) and (ii)
$\frac{f}{f + 6} = \sqrt{\frac{T}{T^{'}}}$
$\frac{600}{606} = \sqrt{\frac{T}{T^{'}}}$
$\frac{T^{'}}{T} = {(\frac{101}{100})}^{2}$
$\frac{T^{'}}{T} - 1 = \frac{2}{100}$
$\frac{T^{'} - T}{T} = 0.02$
$\frac{Δ T}{T} = 0.02$

VITEEE-2009

WAVES

172819 Two sources $A$ and $B$ are sending notes of frequency $680 Hz$ . A listener moves from $A$ and $B$ with a constant velocity $u$ . If the speed of sound in air is $340 {ms}^{- 1}$ , what must be the value of $u$ so that he hears 10 beats per second?

1

2.0 {ms}^{- 1}

2

2.5 {ms}^{- 1}

3

3.0 {ms}^{- 1}

4

3.5 {ms}^{- 1}

Explanation:

B Let listener go from A $\to$ B with velocity (u).

and the apparent frequency of sound from source A by listener using Doppler's effect,
$n^{'} = n (\frac{v - v_{0}}{v + v_{s}}) \Rightarrow n^{'} = 680 (\frac{340 - u}{340 + 0})$
The apparent frequency of sound from source B by listener
$n^{''} = n (\frac{v + v_{0}}{v - v_{s}}) = 680 (\frac{340 + u}{340 - 0})$
Listener hear 10 beats per second.
Hence, $n^{''} - n^{'} = 10$
$\Rightarrow 680 (\frac{340 + u}{340}) - 680 (\frac{340 - u}{340}) = 10$
$\Rightarrow 2 (340 + u - 340 + u) = 10$
$\Rightarrow u = 2.5 {ms}^{- 1}$

VITEEE-2009

WAVES

172814 Wire having tension $225 N$ produces six beats per second when it is tuned with a fork. When tension changes to $256 N$ , it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

1

186 Hz

2

225 Hz

3

256 Hz

4

280 Hz

Explanation:

A For wire, vibrating under tension, the fundamental frequency is given by
$f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
Where, $T$ is tension $& μ$ is mass per unit length of the string.
It is given that, when it is sounded with a tuning fork of frequency $x, 6$ beats per seconds were heard
$∴ (x - f_{1}) = 6$
$f_{1} = x - 6$
$∴ f_{1} = \frac{1}{2 l} \sqrt{\frac{225}{μ}}; f_{2} = \frac{1}{2 l} \sqrt{\frac{256}{μ}}$
$\frac{f_{1}}{f_{2}} = \frac{15}{16}$
$f_{2} = \frac{16}{15} \times f_{1} = \frac{16}{15} (x - 6)$
$But f_{2} - x = 6$
$f_{2} = (x + 6)$
$∴ (x + 6) = \frac{16}{15} (x - 6)$
$15 x + 90 = 16 x - 96$
$x = 186 Hz$

MHT-CET 2016

WAVES

172815 A source of unknown frequency gives 4 beats $s^{- 1}$ when sounded with a source of known frequency $250 Hz$ . The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 Hz$ . The unknown frequency is

1

254 Hz

2

246 Hz

3

240 Hz

4

260 Hz

Explanation:

A $f_{known} = 250 Hz$
$n = 4 beats / sec$
$f_{unknown} = 250 \pm 4$
$= 246 Hz or 254 Hz$
Now, the second harmonic of the source of unknown frequency, $2 f = 513 \pm 5$
$= 513 \pm 5$
$= 518 Hz or 508 Hz$
$∴ I^{st}$ harmonic $= \frac{518}{2}$ or $\frac{508}{2}$
$= 259 or 254$
Hence, the unknown frequency $= 254 Hz$

MHT-CET 2013

WAVES

172817 A note has a frequency $128 Hz$ . The frequency of a note two octaves higher than it is

1

256 Hz

2

64 Hz

3

32 Hz

4

512 Hz

Explanation:

D We know
$n = \log_{2} (\frac{f_{2}}{f_{1}})$
$2 = \log_{2} (\frac{f_{2}}{f_{1}})$
$2^{2} = \frac{f_{2}}{f_{1}}$
$f_{2} = 4 \times f_{1}$
$f_{2} = 4 \times 128$
$f_{2} = 512 Hz = 2$

MHT-CET 2006

WAVES

172818 Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

B We know that,
$f = 600 Hz$
$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
If increase in tension of one wires then beats per second is -
$f + 6 = \frac{1}{2 l} \sqrt{\frac{T^{'}}{m}}$
Dividing the equation (i) and (ii)
$\frac{f}{f + 6} = \sqrt{\frac{T}{T^{'}}}$
$\frac{600}{606} = \sqrt{\frac{T}{T^{'}}}$
$\frac{T^{'}}{T} = {(\frac{101}{100})}^{2}$
$\frac{T^{'}}{T} - 1 = \frac{2}{100}$
$\frac{T^{'} - T}{T} = 0.02$
$\frac{Δ T}{T} = 0.02$

VITEEE-2009

WAVES

172819 Two sources $A$ and $B$ are sending notes of frequency $680 Hz$ . A listener moves from $A$ and $B$ with a constant velocity $u$ . If the speed of sound in air is $340 {ms}^{- 1}$ , what must be the value of $u$ so that he hears 10 beats per second?

1

2.0 {ms}^{- 1}

2

2.5 {ms}^{- 1}

3

3.0 {ms}^{- 1}

4

3.5 {ms}^{- 1}

Explanation:

B Let listener go from A $\to$ B with velocity (u).

and the apparent frequency of sound from source A by listener using Doppler's effect,
$n^{'} = n (\frac{v - v_{0}}{v + v_{s}}) \Rightarrow n^{'} = 680 (\frac{340 - u}{340 + 0})$
The apparent frequency of sound from source B by listener
$n^{''} = n (\frac{v + v_{0}}{v - v_{s}}) = 680 (\frac{340 + u}{340 - 0})$
Listener hear 10 beats per second.
Hence, $n^{''} - n^{'} = 10$
$\Rightarrow 680 (\frac{340 + u}{340}) - 680 (\frac{340 - u}{340}) = 10$
$\Rightarrow 2 (340 + u - 340 + u) = 10$
$\Rightarrow u = 2.5 {ms}^{- 1}$

VITEEE-2009

WAVES

172814 Wire having tension $225 N$ produces six beats per second when it is tuned with a fork. When tension changes to $256 N$ , it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

1

186 Hz

2

225 Hz

3

256 Hz

4

280 Hz

Explanation:

A For wire, vibrating under tension, the fundamental frequency is given by
$f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
Where, $T$ is tension $& μ$ is mass per unit length of the string.
It is given that, when it is sounded with a tuning fork of frequency $x, 6$ beats per seconds were heard
$∴ (x - f_{1}) = 6$
$f_{1} = x - 6$
$∴ f_{1} = \frac{1}{2 l} \sqrt{\frac{225}{μ}}; f_{2} = \frac{1}{2 l} \sqrt{\frac{256}{μ}}$
$\frac{f_{1}}{f_{2}} = \frac{15}{16}$
$f_{2} = \frac{16}{15} \times f_{1} = \frac{16}{15} (x - 6)$
$But f_{2} - x = 6$
$f_{2} = (x + 6)$
$∴ (x + 6) = \frac{16}{15} (x - 6)$
$15 x + 90 = 16 x - 96$
$x = 186 Hz$

MHT-CET 2016

WAVES

172815 A source of unknown frequency gives 4 beats $s^{- 1}$ when sounded with a source of known frequency $250 Hz$ . The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 Hz$ . The unknown frequency is

1

254 Hz

2

246 Hz

3

240 Hz

4

260 Hz

Explanation:

A $f_{known} = 250 Hz$
$n = 4 beats / sec$
$f_{unknown} = 250 \pm 4$
$= 246 Hz or 254 Hz$
Now, the second harmonic of the source of unknown frequency, $2 f = 513 \pm 5$
$= 513 \pm 5$
$= 518 Hz or 508 Hz$
$∴ I^{st}$ harmonic $= \frac{518}{2}$ or $\frac{508}{2}$
$= 259 or 254$
Hence, the unknown frequency $= 254 Hz$

MHT-CET 2013

WAVES

172817 A note has a frequency $128 Hz$ . The frequency of a note two octaves higher than it is

1

256 Hz

2

64 Hz

3

32 Hz

4

512 Hz

Explanation:

D We know
$n = \log_{2} (\frac{f_{2}}{f_{1}})$
$2 = \log_{2} (\frac{f_{2}}{f_{1}})$
$2^{2} = \frac{f_{2}}{f_{1}}$
$f_{2} = 4 \times f_{1}$
$f_{2} = 4 \times 128$
$f_{2} = 512 Hz = 2$

MHT-CET 2006

WAVES

172818 Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

B We know that,
$f = 600 Hz$
$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
If increase in tension of one wires then beats per second is -
$f + 6 = \frac{1}{2 l} \sqrt{\frac{T^{'}}{m}}$
Dividing the equation (i) and (ii)
$\frac{f}{f + 6} = \sqrt{\frac{T}{T^{'}}}$
$\frac{600}{606} = \sqrt{\frac{T}{T^{'}}}$
$\frac{T^{'}}{T} = {(\frac{101}{100})}^{2}$
$\frac{T^{'}}{T} - 1 = \frac{2}{100}$
$\frac{T^{'} - T}{T} = 0.02$
$\frac{Δ T}{T} = 0.02$

VITEEE-2009

WAVES

172819 Two sources $A$ and $B$ are sending notes of frequency $680 Hz$ . A listener moves from $A$ and $B$ with a constant velocity $u$ . If the speed of sound in air is $340 {ms}^{- 1}$ , what must be the value of $u$ so that he hears 10 beats per second?

1

2.0 {ms}^{- 1}

2

2.5 {ms}^{- 1}

3

3.0 {ms}^{- 1}

4

3.5 {ms}^{- 1}

Explanation:

B Let listener go from A $\to$ B with velocity (u).

and the apparent frequency of sound from source A by listener using Doppler's effect,
$n^{'} = n (\frac{v - v_{0}}{v + v_{s}}) \Rightarrow n^{'} = 680 (\frac{340 - u}{340 + 0})$
The apparent frequency of sound from source B by listener
$n^{''} = n (\frac{v + v_{0}}{v - v_{s}}) = 680 (\frac{340 + u}{340 - 0})$
Listener hear 10 beats per second.
Hence, $n^{''} - n^{'} = 10$
$\Rightarrow 680 (\frac{340 + u}{340}) - 680 (\frac{340 - u}{340}) = 10$
$\Rightarrow 2 (340 + u - 340 + u) = 10$
$\Rightarrow u = 2.5 {ms}^{- 1}$

VITEEE-2009

WAVES

172814 Wire having tension $225 N$ produces six beats per second when it is tuned with a fork. When tension changes to $256 N$ , it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

1

186 Hz

2

225 Hz

3

256 Hz

4

280 Hz

Explanation:

A For wire, vibrating under tension, the fundamental frequency is given by
$f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
Where, $T$ is tension $& μ$ is mass per unit length of the string.
It is given that, when it is sounded with a tuning fork of frequency $x, 6$ beats per seconds were heard
$∴ (x - f_{1}) = 6$
$f_{1} = x - 6$
$∴ f_{1} = \frac{1}{2 l} \sqrt{\frac{225}{μ}}; f_{2} = \frac{1}{2 l} \sqrt{\frac{256}{μ}}$
$\frac{f_{1}}{f_{2}} = \frac{15}{16}$
$f_{2} = \frac{16}{15} \times f_{1} = \frac{16}{15} (x - 6)$
$But f_{2} - x = 6$
$f_{2} = (x + 6)$
$∴ (x + 6) = \frac{16}{15} (x - 6)$
$15 x + 90 = 16 x - 96$
$x = 186 Hz$

MHT-CET 2016

WAVES

172815 A source of unknown frequency gives 4 beats $s^{- 1}$ when sounded with a source of known frequency $250 Hz$ . The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 Hz$ . The unknown frequency is

1

254 Hz

2

246 Hz

3

240 Hz

4

260 Hz

Explanation:

A $f_{known} = 250 Hz$
$n = 4 beats / sec$
$f_{unknown} = 250 \pm 4$
$= 246 Hz or 254 Hz$
Now, the second harmonic of the source of unknown frequency, $2 f = 513 \pm 5$
$= 513 \pm 5$
$= 518 Hz or 508 Hz$
$∴ I^{st}$ harmonic $= \frac{518}{2}$ or $\frac{508}{2}$
$= 259 or 254$
Hence, the unknown frequency $= 254 Hz$

MHT-CET 2013

WAVES

172817 A note has a frequency $128 Hz$ . The frequency of a note two octaves higher than it is

1

256 Hz

2

64 Hz

3

32 Hz

4

512 Hz

Explanation:

D We know
$n = \log_{2} (\frac{f_{2}}{f_{1}})$
$2 = \log_{2} (\frac{f_{2}}{f_{1}})$
$2^{2} = \frac{f_{2}}{f_{1}}$
$f_{2} = 4 \times f_{1}$
$f_{2} = 4 \times 128$
$f_{2} = 512 Hz = 2$

MHT-CET 2006

WAVES

172818 Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

B We know that,
$f = 600 Hz$
$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
If increase in tension of one wires then beats per second is -
$f + 6 = \frac{1}{2 l} \sqrt{\frac{T^{'}}{m}}$
Dividing the equation (i) and (ii)
$\frac{f}{f + 6} = \sqrt{\frac{T}{T^{'}}}$
$\frac{600}{606} = \sqrt{\frac{T}{T^{'}}}$
$\frac{T^{'}}{T} = {(\frac{101}{100})}^{2}$
$\frac{T^{'}}{T} - 1 = \frac{2}{100}$
$\frac{T^{'} - T}{T} = 0.02$
$\frac{Δ T}{T} = 0.02$

VITEEE-2009

WAVES

172819 Two sources $A$ and $B$ are sending notes of frequency $680 Hz$ . A listener moves from $A$ and $B$ with a constant velocity $u$ . If the speed of sound in air is $340 {ms}^{- 1}$ , what must be the value of $u$ so that he hears 10 beats per second?

1

2.0 {ms}^{- 1}

2

2.5 {ms}^{- 1}

3

3.0 {ms}^{- 1}

4

3.5 {ms}^{- 1}

Explanation:

B Let listener go from A $\to$ B with velocity (u).

and the apparent frequency of sound from source A by listener using Doppler's effect,
$n^{'} = n (\frac{v - v_{0}}{v + v_{s}}) \Rightarrow n^{'} = 680 (\frac{340 - u}{340 + 0})$
The apparent frequency of sound from source B by listener
$n^{''} = n (\frac{v + v_{0}}{v - v_{s}}) = 680 (\frac{340 + u}{340 - 0})$
Listener hear 10 beats per second.
Hence, $n^{''} - n^{'} = 10$
$\Rightarrow 680 (\frac{340 + u}{340}) - 680 (\frac{340 - u}{340}) = 10$
$\Rightarrow 2 (340 + u - 340 + u) = 10$
$\Rightarrow u = 2.5 {ms}^{- 1}$

VITEEE-2009

WAVES

172814 Wire having tension $225 N$ produces six beats per second when it is tuned with a fork. When tension changes to $256 N$ , it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

1

186 Hz

2

225 Hz

3

256 Hz

4

280 Hz

Explanation:

A For wire, vibrating under tension, the fundamental frequency is given by
$f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
Where, $T$ is tension $& μ$ is mass per unit length of the string.
It is given that, when it is sounded with a tuning fork of frequency $x, 6$ beats per seconds were heard
$∴ (x - f_{1}) = 6$
$f_{1} = x - 6$
$∴ f_{1} = \frac{1}{2 l} \sqrt{\frac{225}{μ}}; f_{2} = \frac{1}{2 l} \sqrt{\frac{256}{μ}}$
$\frac{f_{1}}{f_{2}} = \frac{15}{16}$
$f_{2} = \frac{16}{15} \times f_{1} = \frac{16}{15} (x - 6)$
$But f_{2} - x = 6$
$f_{2} = (x + 6)$
$∴ (x + 6) = \frac{16}{15} (x - 6)$
$15 x + 90 = 16 x - 96$
$x = 186 Hz$

MHT-CET 2016

WAVES

172815 A source of unknown frequency gives 4 beats $s^{- 1}$ when sounded with a source of known frequency $250 Hz$ . The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 Hz$ . The unknown frequency is

1

254 Hz

2

246 Hz

3

240 Hz

4

260 Hz

Explanation:

A $f_{known} = 250 Hz$
$n = 4 beats / sec$
$f_{unknown} = 250 \pm 4$
$= 246 Hz or 254 Hz$
Now, the second harmonic of the source of unknown frequency, $2 f = 513 \pm 5$
$= 513 \pm 5$
$= 518 Hz or 508 Hz$
$∴ I^{st}$ harmonic $= \frac{518}{2}$ or $\frac{508}{2}$
$= 259 or 254$
Hence, the unknown frequency $= 254 Hz$

MHT-CET 2013

WAVES

172817 A note has a frequency $128 Hz$ . The frequency of a note two octaves higher than it is

1

256 Hz

2

64 Hz

3

32 Hz

4

512 Hz

Explanation:

D We know
$n = \log_{2} (\frac{f_{2}}{f_{1}})$
$2 = \log_{2} (\frac{f_{2}}{f_{1}})$
$2^{2} = \frac{f_{2}}{f_{1}}$
$f_{2} = 4 \times f_{1}$
$f_{2} = 4 \times 128$
$f_{2} = 512 Hz = 2$

MHT-CET 2006

WAVES

172818 Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

B We know that,
$f = 600 Hz$
$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
If increase in tension of one wires then beats per second is -
$f + 6 = \frac{1}{2 l} \sqrt{\frac{T^{'}}{m}}$
Dividing the equation (i) and (ii)
$\frac{f}{f + 6} = \sqrt{\frac{T}{T^{'}}}$
$\frac{600}{606} = \sqrt{\frac{T}{T^{'}}}$
$\frac{T^{'}}{T} = {(\frac{101}{100})}^{2}$
$\frac{T^{'}}{T} - 1 = \frac{2}{100}$
$\frac{T^{'} - T}{T} = 0.02$
$\frac{Δ T}{T} = 0.02$

VITEEE-2009

WAVES

172819 Two sources $A$ and $B$ are sending notes of frequency $680 Hz$ . A listener moves from $A$ and $B$ with a constant velocity $u$ . If the speed of sound in air is $340 {ms}^{- 1}$ , what must be the value of $u$ so that he hears 10 beats per second?

1

2.0 {ms}^{- 1}

2

2.5 {ms}^{- 1}

3

3.0 {ms}^{- 1}

4

3.5 {ms}^{- 1}

Explanation:

B Let listener go from A $\to$ B with velocity (u).

and the apparent frequency of sound from source A by listener using Doppler's effect,
$n^{'} = n (\frac{v - v_{0}}{v + v_{s}}) \Rightarrow n^{'} = 680 (\frac{340 - u}{340 + 0})$
The apparent frequency of sound from source B by listener
$n^{''} = n (\frac{v + v_{0}}{v - v_{s}}) = 680 (\frac{340 + u}{340 - 0})$
Listener hear 10 beats per second.
Hence, $n^{''} - n^{'} = 10$
$\Rightarrow 680 (\frac{340 + u}{340}) - 680 (\frac{340 - u}{340}) = 10$
$\Rightarrow 2 (340 + u - 340 + u) = 10$
$\Rightarrow u = 2.5 {ms}^{- 1}$

VITEEE-2009

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