A particle moves in a circle of radius $5cm$ with constant speed and time period $0.2\pi s$. The acceleration of the particle is
A. $0\dfrac{m}{{{s^2}}}$
B. $36\dfrac{m}{{{s^2}}}$
C. $5\dfrac{m}{{{s^2}}}$
D. $15\dfrac{m}{{{s^2}}}$
Answer
641.7k+ views
Hint: velocity can be defined as the rate of displacement. And the acceleration is the rate of velocity. If the velocity changes by time, it is called acceleration.
Formula used:
(i) \[V = \dfrac{{2\pi r}}{t}\]
(ii) \[a = \dfrac{{{V^2}}}{r}\]
Where,
\[V\]=velocity
\[r\]=radius
\[t\]=time period
\[a\]=acceleration
Complete step by step answer:
To know the acceleration, we have to know what the velocity is. Therefore to find the velocity we use the formula
\[V = \dfrac{{2\pi r}}{t}\]
From the question, we have radius, \[r = 5cm\]. We can convert the centimetres into meters we get \[r\]\[ = 5 \times {10^{ - 2}}m\] and the time period, \[t = 0.2\pi s\]. Apply these values in the above velocity formula.
\[V = \dfrac{{2 \times 3.14 \times 5 \times {{10}^{ - 2}}}}{{0.2 \times 3.14}}\]
Now let us cancel the common terms, then we have
\[V = \dfrac{{5 \times {{10}^{ - 2}}}}{{0.1}}\]
\[ \Rightarrow 5 \times {10^{ - 2 + 1}}\]
\[V = 0.5m{s^{ - 1}}\]
Now we are going to apply the value of velocity in the acceleration formula. As the particle is moving in the circle, it experiences centripetal force. The acceleration is given as the velocity divided by the distance. Hence the acceleration is given as,
\[a = \dfrac{{{V^2}}}{r}\]
Apply the known values in the formula,
\[a = \dfrac{{{{\left( {0.5} \right)}^2}}}{{5 \times {{10}^{ - 2}}}}\]
\[ \Rightarrow \dfrac{{0.25}}{{0.05}}\]
\[ \Rightarrow 5\]
Hence, the acceleration \[a = 5m{s^{ - 2}}\]
Therefore the correct option is option C.
Additional information:
(i) Velocity is the speed taken by a particle to reach a specific distance in a specific time. Its unit is \[m{s^{ - 1}}\]. Acceleration is the change of velocities while reaching the specific distance in the specific time. Its unit is \[m{s^{ - 2}}\]
(ii)The dimension of velocity is \[L{T^{ - 1}}\]. And the dimensional formula for acceleration is \[L{T^{ - 2}}\].
(iii)Velocity and acceleration are the vector quantities. Which means the value denotes both magnitude and direction.
Note:
While calculating the acceleration and velocity, it is very important to know about the path it travels like the shape of the path.
Formula used:
(i) \[V = \dfrac{{2\pi r}}{t}\]
(ii) \[a = \dfrac{{{V^2}}}{r}\]
Where,
\[V\]=velocity
\[r\]=radius
\[t\]=time period
\[a\]=acceleration
Complete step by step answer:
To know the acceleration, we have to know what the velocity is. Therefore to find the velocity we use the formula
\[V = \dfrac{{2\pi r}}{t}\]
From the question, we have radius, \[r = 5cm\]. We can convert the centimetres into meters we get \[r\]\[ = 5 \times {10^{ - 2}}m\] and the time period, \[t = 0.2\pi s\]. Apply these values in the above velocity formula.
\[V = \dfrac{{2 \times 3.14 \times 5 \times {{10}^{ - 2}}}}{{0.2 \times 3.14}}\]
Now let us cancel the common terms, then we have
\[V = \dfrac{{5 \times {{10}^{ - 2}}}}{{0.1}}\]
\[ \Rightarrow 5 \times {10^{ - 2 + 1}}\]
\[V = 0.5m{s^{ - 1}}\]
Now we are going to apply the value of velocity in the acceleration formula. As the particle is moving in the circle, it experiences centripetal force. The acceleration is given as the velocity divided by the distance. Hence the acceleration is given as,
\[a = \dfrac{{{V^2}}}{r}\]
Apply the known values in the formula,
\[a = \dfrac{{{{\left( {0.5} \right)}^2}}}{{5 \times {{10}^{ - 2}}}}\]
\[ \Rightarrow \dfrac{{0.25}}{{0.05}}\]
\[ \Rightarrow 5\]
Hence, the acceleration \[a = 5m{s^{ - 2}}\]
Therefore the correct option is option C.
Additional information:
(i) Velocity is the speed taken by a particle to reach a specific distance in a specific time. Its unit is \[m{s^{ - 1}}\]. Acceleration is the change of velocities while reaching the specific distance in the specific time. Its unit is \[m{s^{ - 2}}\]
(ii)The dimension of velocity is \[L{T^{ - 1}}\]. And the dimensional formula for acceleration is \[L{T^{ - 2}}\].
(iii)Velocity and acceleration are the vector quantities. Which means the value denotes both magnitude and direction.
Note:
While calculating the acceleration and velocity, it is very important to know about the path it travels like the shape of the path.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

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

10 examples of friction in our daily life

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

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

