- #1
WMDhamnekar
MHB
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- Homework Statement
- Q.2:The two charges of ##-4\mu C## and ##+ 4\mu C ##are placed at the point A(1,0.4)m and B( 2,-1,5)m located in an electric field ##\overrightarrow{E}=0.20 \hat{i} V/cm.##The magnitude of Torque acting on the dipole is ##8\sqrt{\alpha} \times 10^{-5} Nm ## What is the value of ##\alpha?##
Q.3 :A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. What is the velocity of sound in meter per second?
- Relevant Equations
- Not applicable
Relevant video ##\Rightarrow## Physics questions and answers
My answer to Q.2
4. Compare with the given magnitude:We are given that the magnitude of the torque is ##8\sqrt{\alpha} \times 10^{-5} N⋅m.##So, ##80\sqrt{}6 \times 10^{-6} = 8\sqrt{\alpha} \times 10^{-5}##
Simplifying, we get:##\sqrt{6} = \sqrt{\alpha}##
Therefore, ##\alpha =6 ## But author of video said ##\alpha =2##
My answer to Q.3
Let's denote the frequency of the closed organ pipe as ##f_1## and the frequency of the open organ pipe as ##f_2.##
Fundamental Frequency of a Closed Organ Pipe: ##f_1 = v / 4L_1 ##where:
Substituting the expressions for ## f_1## and ##f_2:##
##(v / 4L_1) - (v / 2L_2) = 7##
##v(1/4L_1 - 1/2L_2) = 7##
##v[(2L_2 - 4L_1) / (4L_1L_2)] = 7##
##v = 7 * (4L_1L_2) / (2L_2 - 4L_1)##
Substituting the given values:
v = 7 * (4 * 1.5 * 3.5) / (2 * 3.5 - 4 * 1.5)
v = 7 * 21 / 1
v = 147 m/s
Therefore, the velocity of sound in air is 147 m/s. But the author said velocity of sound is 294 m/s. How is that?
My answer to Q.2
Let's break down the problem and solve it step-by-step:
1. Find the dipole moment (p):- The dipole moment is defined as the product of the charge (q) and the separation distance (d) between the charges.
- p = qd
- Here, ##q = 4 \mu C = 4 × 10^{-6} C##
- To find the separation distance, we need to find the vector difference between the positions of the two charges:##d = B - A = (2 - 1)\hat{i} + (-1 - 0)\hat{j} + (5 - 4)k\hat{k} = \hat{i} - \hat{j} + \hat{k}##
- The magnitude of d is ## |d| = \sqrt(1^2 + (-1)^2 + 1^2) = \sqrt{3} m##
- So, the dipole moment ##\rho = (4 × 10^{-6} C) \times (\sqrt{3} m) = 4\sqrt{3} \times 10^{-6} C⋅m##
- The torque on a dipole in a uniform electric field is given by:##\tau = \rho \times E ##
- Here, ##E = 0.20 \hat{i} V/cm = 20 \hat{i} V/m##
- So, ##\tau = (4\sqrt{3} \times 10^{-6} C⋅m) \times (20 \hat{i} V/m)##
- Calculating the cross product, we get:##\tau = 80\sqrt{3} \times 10^{-6} (\hat{j} + \hat{k}) N⋅m##
4. Compare with the given magnitude:We are given that the magnitude of the torque is ##8\sqrt{\alpha} \times 10^{-5} N⋅m.##So, ##80\sqrt{}6 \times 10^{-6} = 8\sqrt{\alpha} \times 10^{-5}##
Simplifying, we get:##\sqrt{6} = \sqrt{\alpha}##
Therefore, ##\alpha =6 ## But author of video said ##\alpha =2##
My answer to Q.3
Let's denote the frequency of the closed organ pipe as ##f_1## and the frequency of the open organ pipe as ##f_2.##
Fundamental Frequency of a Closed Organ Pipe: ##f_1 = v / 4L_1 ##where:
- ##f_1## is the fundamental frequency
- v is the velocity of sound
- ##L_1## is the length of the closed pipe
- ##f_2## is the fundamental frequency
- v is the velocity of sound
- ##L_2## is the length of the open pipe
- ##L_1 = 150 cm = 1.5 m##
- ##L_2 = 350 cm = 3.5 m##
- Beat frequency = 7 Hz (This means ##|f_1 - f_2| = 7)##
Substituting the expressions for ## f_1## and ##f_2:##
##(v / 4L_1) - (v / 2L_2) = 7##
##v(1/4L_1 - 1/2L_2) = 7##
##v[(2L_2 - 4L_1) / (4L_1L_2)] = 7##
##v = 7 * (4L_1L_2) / (2L_2 - 4L_1)##
Substituting the given values:
v = 7 * (4 * 1.5 * 3.5) / (2 * 3.5 - 4 * 1.5)
v = 7 * 21 / 1
v = 147 m/s
Therefore, the velocity of sound in air is 147 m/s. But the author said velocity of sound is 294 m/s. How is that?
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