On a new scale of temperature (which is linea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The relationship between any two linear temperature scales is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{freezing}} = \frac{C

Vedclass · Accepted Answer

$117$

Vedclass · Answer

(D) The relationship between any two linear temperature scales is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{freezing}} = \frac{C - C_{freezing}}{C_{boiling} - C_{freezing}}$.For the $W$ scale,the freezing point is $39\,^{\circ}W$ and the boiling point is $239\,^{\circ}W$.For the Celsius scale,the freezing point is $0\,^{\circ}C$ and the boiling point is $100\,^{\circ}C$.Substituting the values into the formula:$\frac{W - 39}{239 - 39} = \frac{39 - 0}{100 - 0}$$\frac{W - 39}{200} = \frac{39}{100}$$W - 39 = \frac{39 	imes 200}{100}$$W - 39 = 39 	imes 2 = 78$$W = 78 + 39 = 117\,^{\circ}W$.

Similar Questions

$A$ thin metallic spherical shell contains a charge $Q$ on it. $A$ point charge $+q$ is placed at the centre of the shell and another charge $q'$ is placed outside it as shown in the figure. All the three charges are positive. The force on the charge at the centre is:

Given that $\Delta H_f(H) = 218 \ kJ/mol$,express the $H-H$ bond energy in $kcal/mol$.

The angle of a prism is $A.$ One of its refracting surfaces is silvered. Light rays falling at an angle of incidence $2A$ on the first surface return back through the same path after suffering reflection at the silvered surface. The refractive index $\mu$ of the prism is

Suppose that a random variable $X$ follows a Poisson distribution. If $P(X=1) = P(X=2)$,then $P(X=5)$ is equal to:

The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$,where the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module