
An aeroplane having a tip to tip distance of its wings as 50m. if the vertical components of the magnetic field of the earth is $4\times {{10}^{-5}}T$ then the potential difference developed across the wing is 0.2V. The speed of the plane is:
$\begin{align}
& A.\text{ }100Km/hr \\
& B.\text{ }360Km/hr \\
& C.\text{ }720Km/hr \\
& D.\text{ }180Km/hr \\
\end{align}$
Answer
498.3k+ views
Hint: In order to solve this question we can use formula for the induced emf due to the magnetic field of the earth which is given in the question that the emf or the potential difference developed across the wing therefore we can use the formula for the induced emf to find the velocity.
Formula used:
$\varepsilon =Bvl$
Complete answer:
It is given in the question that the emf induced across the wing is 0.2V now we know that formula for induced emf due to the magnetic field.
$\varepsilon =Bvl...\left( 1 \right)$
Where, ε = induced emf across wing
B = magnetic field
v = velocity
l = length of the object
Now here value of the magnetic field is given as,
$B=4\times {{10}^{-5}}T$
And we can take distance between tip to wings as length of an object because emf is developed across the wing of an aeroplane therefore,
l = 50m
Now substitute all the values in the equation (1)
$\begin{align}
& \Rightarrow 0.2=4\times {{10}^{-5}}\times v\times 50 \\
& \Rightarrow v=\dfrac{0.2}{200\times {{10}^{-5}}} \\
& \Rightarrow v=\dfrac{0.2}{0.2\times {{10}^{-5}}} \\
& \therefore v=100m/s \\
\end{align}$
Now let’s convert units from the m/s to Km/hr to get relevant answer
$\begin{align}
& v=100\times \dfrac{18}{5} \\
& v=360Km/h \\
\end{align}$
Therefore option (B) 360Km/h is correct.
Note:
As we can see that when we find velocity of the aeroplane is 100m/s but if we forgot to change the units into Km/h then we can easily mistake that option (A) which is given as 100Km/h is correct but it is eventually wrong as we can see that after converting units from m/s to Km/h the answer is 360Km/h which is the correct answer.
Formula used:
$\varepsilon =Bvl$
Complete answer:
It is given in the question that the emf induced across the wing is 0.2V now we know that formula for induced emf due to the magnetic field.
$\varepsilon =Bvl...\left( 1 \right)$
Where, ε = induced emf across wing
B = magnetic field
v = velocity
l = length of the object
Now here value of the magnetic field is given as,
$B=4\times {{10}^{-5}}T$
And we can take distance between tip to wings as length of an object because emf is developed across the wing of an aeroplane therefore,
l = 50m
Now substitute all the values in the equation (1)
$\begin{align}
& \Rightarrow 0.2=4\times {{10}^{-5}}\times v\times 50 \\
& \Rightarrow v=\dfrac{0.2}{200\times {{10}^{-5}}} \\
& \Rightarrow v=\dfrac{0.2}{0.2\times {{10}^{-5}}} \\
& \therefore v=100m/s \\
\end{align}$
Now let’s convert units from the m/s to Km/hr to get relevant answer
$\begin{align}
& v=100\times \dfrac{18}{5} \\
& v=360Km/h \\
\end{align}$
Therefore option (B) 360Km/h is correct.
Note:
As we can see that when we find velocity of the aeroplane is 100m/s but if we forgot to change the units into Km/h then we can easily mistake that option (A) which is given as 100Km/h is correct but it is eventually wrong as we can see that after converting units from m/s to Km/h the answer is 360Km/h which is the correct answer.
Recently Updated Pages
Full Form: Updated List (2025) of Important Abbreviations You Should Know

CAD Full Form: Meaning in Engineering, Medical & Banking

Counting Money Worksheet for Class 3 Maths – Free PDF

COB Full Form: Meaning, Uses & Examples for Students

BSc Full Form: Bachelor of Science Degree, Scope & Courses

7 Times Table Worksheet PDF: Free Printable Math Drills

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
