Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q . The relationship for conversion between the two scales is given   by :-
 
 

Option: 1

\frac{t_{p}}{100}= \frac{t_{Q}-180}{150}


Option: 2

\frac{t_Q}{150}=\frac{t_P-180}{100}


Option: 3

\frac{t_p}{180}-\frac{t_Q-40}{100}


Option: 4

\frac{t_Q}{100}=\frac{t_P-30}{150}


Answers (1)

best_answer

   
\begin{aligned} & \frac{\mathrm{t}_{\mathrm{p}}-30}{180-30}=\frac{\mathrm{t}_{\mathrm{Q}}-0}{100-0} \\ & \frac{\mathrm{t}_{\mathrm{p}}-30}{150}=\frac{\mathrm{t}_{\mathrm{Q}}}{100} \\ & \frac{\mathrm{t}_{\mathrm{Q}}}{100}=\frac{\mathrm{t}_{\mathrm{P}}-30}{150} \end{aligned}

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BSEB Super 50 performed ‘brilliantly’ in JEE Main, 23 scored above 95 percentile: Bihar Board chairman
anam-khan

IIIT Allahabad cut-offs for BTech ECE, IT, IT-Business Informatics goes up in last 3 years; seat matrix
anam-khan

IIIT Hyderabad JEE Main cut-offs for BTech CSE, ECE of last 5 years; modes of admissions for dual degrees

Latest Question