
The ratio of the radius of the back wheel of a tractor to that of the front wheel is \[7:5\] . the diameter of the back wheel is \[70cm\] . what is the distance moved by the front wheel after it has turned \[10\] rounds? (take \[\pi =\dfrac{22}{7}\] round off your answer to the nearest whole number)
A) \[1440cm\]
B) \[1500cm\]
C) \[1571cm\]
D) \[1600cm\]
Answer
509.1k+ views
Hint: From the given data, to find the distance moved by the front wheel we need to find the radius of the front wheel by using the radius of the back wheel and ratio given. Then the circumference of the front wheel is to be found by using formula \[C=2\pi r\] . and finally multiply the result with ten rounds.
Complete step-by-step solution:
Let the radius of front wheel is given by \[{{r}_{F}}\]
And the radius of back wheel is given by \[{{r}_{B}}\]
Given ratio is
\[{{r}_{B}}:{{r}_{F}}=7:5\]
The diameter of back wheel is given
$ {{d}_{B}}=70cm $
$ \Rightarrow {{r}_{B}}=\dfrac{{{d}_{B}}}{2}=\dfrac{70}{2}=35cm $
Substitute the value in above equation
$ \Rightarrow 35:{{r}_{F}}=7:5 $
$ \Rightarrow \dfrac{35}{{{r}_{F}}}=\dfrac{7}{5} $
$ \Rightarrow {{r}_{F}}=\dfrac{5}{7}\times 35 $
$ \Rightarrow {{r}_{F}}=25cm $
The circumference of the tractor’s front wheel is
$ C=2\pi {{r}_{F}} $
$ \Rightarrow C=2\pi \times 25 $
$ \Rightarrow C=50\pi cm $
For one round the distance covered is equal to circumference Since the front wheel has turned \[10\] rounds it means the circumference is multiplied with \[10\] .
Let the total distance covered be \[D\]
We have
$ D=10\times C $
$ \Rightarrow D=10\times 50\pi cm $
$ \Rightarrow D=500\pi cm $
Substitute the value of \[\pi =\dfrac{22}{7}\]. We get
$ \Rightarrow D=500\times \dfrac{22}{7} $
$ \Rightarrow D=500\times \dfrac{22}{7} $
$ \Rightarrow D=\text{71}\text{.428}\times 22 $
$ \Rightarrow D=\text{1,571}\text{.416cm} $
$ \Rightarrow D\approx 1571cm $
Hence the distance covered by the front wheel is nearly equal to \[1571cm\] . It is given by option (C).
Note: If we consider a point on the wheel, as the wheel rotates the point also moves from its initial position to different position. The distance taken to return to the same position of the point is given by the circumference of the given wheel or circle. The circumference of the given circle is given by \[C=2\pi r\] .
Complete step-by-step solution:
Let the radius of front wheel is given by \[{{r}_{F}}\]
And the radius of back wheel is given by \[{{r}_{B}}\]
Given ratio is
\[{{r}_{B}}:{{r}_{F}}=7:5\]
The diameter of back wheel is given
$ {{d}_{B}}=70cm $
$ \Rightarrow {{r}_{B}}=\dfrac{{{d}_{B}}}{2}=\dfrac{70}{2}=35cm $
Substitute the value in above equation
$ \Rightarrow 35:{{r}_{F}}=7:5 $
$ \Rightarrow \dfrac{35}{{{r}_{F}}}=\dfrac{7}{5} $
$ \Rightarrow {{r}_{F}}=\dfrac{5}{7}\times 35 $
$ \Rightarrow {{r}_{F}}=25cm $
The circumference of the tractor’s front wheel is
$ C=2\pi {{r}_{F}} $
$ \Rightarrow C=2\pi \times 25 $
$ \Rightarrow C=50\pi cm $
For one round the distance covered is equal to circumference Since the front wheel has turned \[10\] rounds it means the circumference is multiplied with \[10\] .
Let the total distance covered be \[D\]
We have
$ D=10\times C $
$ \Rightarrow D=10\times 50\pi cm $
$ \Rightarrow D=500\pi cm $
Substitute the value of \[\pi =\dfrac{22}{7}\]. We get
$ \Rightarrow D=500\times \dfrac{22}{7} $
$ \Rightarrow D=500\times \dfrac{22}{7} $
$ \Rightarrow D=\text{71}\text{.428}\times 22 $
$ \Rightarrow D=\text{1,571}\text{.416cm} $
$ \Rightarrow D\approx 1571cm $
Hence the distance covered by the front wheel is nearly equal to \[1571cm\] . It is given by option (C).
Note: If we consider a point on the wheel, as the wheel rotates the point also moves from its initial position to different position. The distance taken to return to the same position of the point is given by the circumference of the given wheel or circle. The circumference of the given circle is given by \[C=2\pi r\] .
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