Login/Sign Up
The diameters of front and rear wheels of a tractor are \[80\]cm and \[2\] m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes \[1400\] revolutions.
The diameters of front and rear wheels of a tractor are \[80\]cm and \[2\] m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes \[1400\] revolutions.
Areas Related to Circles | Exercise 11.4 | Maths | NCERT Exemplar
The diameters of front and rear wheels of a tractor are \[80\]cm and \[2\] m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes \[1400\] revolutions.

solution

From the given question,
We got, Diameter of front wheels = \[{{d}_{1}}\]= \[80\] cm
we got, Diameter of rear wheels = \[{{d}_{2}}\]= \[2\]m = \[200\] cm
Let us consider \[{{r}_{1}}\] be the radius of the front wheels = \[80/2\] = \[40\] cm
Let us consider \[{{r}_{2}}\] be the radius of the rear wheels = \[200/2\]= \[100\] cm
Now, we know that the Circumference of the front wheels = \[2\pi r\]
= \[2\times (22/7)\times 40\]
= \[1760/7\] cm
Here, Circumference of the rear wheels = \[2\pi r\] = \[2\times (22/7)\times 100\]= \[4400/7\] cm
According to the question, No. of revolutions made by the front wheel = \[1400\]
∴ Distance covered by the front wheel = \[1400\times (1760/7)\]= \[352000\]cm
number of revolutions made by rear wheel is
=distance covered by front wheel/circumference of the rear wheel
= \[\frac{352000}{\frac{4400}{7}}\]
= \[\frac{352000\times 7}{4400}\]
= 560 revolutions.

i

More Material

(i) Consider a thin lens placed between a source (S) and an observer (O). Let the thickness of the lens vary as 2 0 () – α = b wb w, where b is the verticle distance from the pole. w0 is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.
(ii) A gravitational lens may be assumed to have a varying width of the form w(b) = k1 ln (k2/b) = k1 ln (k2/bmin). Show that an observer will see an image of a point object as a ring about the centre of the lens with an angular radius.
β = √(n-1)k1 u/v / u + v

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by n(r) = 1 + 2 GM/rc2 where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.