Answer
Verified
360.6k+ views
Hint: Here, we are given the velocities of both the ends of the link. We know that the link is rigid, and hence the length of the link will be fixed. Thus the resultant velocity of both the components along the link will be the same. Hence, from the given velocity, we can find the velocity along the link and from that, we can find the velocity of the required component.
Complete step by step answer:
Here, we are given a rigid link, whose length remains constant. The link is moving at $60^ \circ$ with the horizontal, with both the ends having velocity of different magnitude and directions. Here, we are given the velocity of the point $A$ as $v_A=3ms^{-1}$.
Now, this is the velocity of the point itself. We can find the velocity of the point along the link $\;AB$ by considering the component of the velocity in the direction of the link as shown in figure.
Hence, the velocity of the point $A$ along the link as shown in the figure is given as
$v'={{v}_{A}}\cos 60{}^\circ $
Substituting the values,
$\therefore v'=\dfrac{3}{2}m{{s}^{-1}}$…… $(1)$
Now, as the link is rigid, the velocity of the point $B$ along the link $\;AB$ will be equal to the velocity of the point $A$ along the link.
From the above figure, the velocity of point $B$ along the link is given as
$v'={{v}_{B}}\cos 30{}^\circ $…… $(2)$
Equating both the equations,
$\therefore \dfrac{3}{2}={{v}_{B}}\times \dfrac{\sqrt{3}}{2}$
$\therefore {{v}_{B}}=\sqrt{3}m{{s}^{-1}}$
Hence, the correct answer is option C.
Note: We can also solve the given question, by assuming links for the given velocities. We know that the velocity is always perpendicular to the link. Hence, we can draw the links for the given velocities, we can draw the links as shown below,
From the figure, we can find the ratio of the links $\cot 60{}^\circ =\dfrac{OB}{OA}$ . Now, we know that $\dfrac{OB}{OA}=\dfrac{{{v}_{B}}}{{{v}_{A}}}$
Hence, we can get the formula as ${{v}_{B}}=\cot 60{}^\circ \times {{v}_{A}}$
$\therefore {{v}_{B}}=\sqrt{3}m{{s}^{-1}}$
Complete step by step answer:
Here, we are given a rigid link, whose length remains constant. The link is moving at $60^ \circ$ with the horizontal, with both the ends having velocity of different magnitude and directions. Here, we are given the velocity of the point $A$ as $v_A=3ms^{-1}$.
Now, this is the velocity of the point itself. We can find the velocity of the point along the link $\;AB$ by considering the component of the velocity in the direction of the link as shown in figure.
Hence, the velocity of the point $A$ along the link as shown in the figure is given as
$v'={{v}_{A}}\cos 60{}^\circ $
Substituting the values,
$\therefore v'=\dfrac{3}{2}m{{s}^{-1}}$…… $(1)$
Now, as the link is rigid, the velocity of the point $B$ along the link $\;AB$ will be equal to the velocity of the point $A$ along the link.
From the above figure, the velocity of point $B$ along the link is given as
$v'={{v}_{B}}\cos 30{}^\circ $…… $(2)$
Equating both the equations,
$\therefore \dfrac{3}{2}={{v}_{B}}\times \dfrac{\sqrt{3}}{2}$
$\therefore {{v}_{B}}=\sqrt{3}m{{s}^{-1}}$
Hence, the correct answer is option C.
Note: We can also solve the given question, by assuming links for the given velocities. We know that the velocity is always perpendicular to the link. Hence, we can draw the links for the given velocities, we can draw the links as shown below,
From the figure, we can find the ratio of the links $\cot 60{}^\circ =\dfrac{OB}{OA}$ . Now, we know that $\dfrac{OB}{OA}=\dfrac{{{v}_{B}}}{{{v}_{A}}}$
Hence, we can get the formula as ${{v}_{B}}=\cot 60{}^\circ \times {{v}_{A}}$
$\therefore {{v}_{B}}=\sqrt{3}m{{s}^{-1}}$
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Advantages and disadvantages of science
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference Between Plant Cell and Animal Cell
10 examples of evaporation in daily life with explanations
Give 10 examples for herbs , shrubs , climbers , creepers
Write a letter to the principal requesting him to grant class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE