MCQ
$-122^{\circ} C$ in Fahrenheit scale is
- A$237.6^{\circ} F$
- B$-317.6^{\circ} F$
- C$317.6^{\circ} F$
- ✓$-187.6^{\circ} F$
✓
Answer
Correct option: D.
$-187.6^{\circ} F$
(d) $-187.6^{\circ} F$
Explanation: Substituting the given value for ${ }^{\circ} C =(-122)$ in the following expression,
$
\begin{aligned}
& { }^{\circ} F=\left[1.8\left({ }^{\circ} C\right)+32\right] \\
& =[1.8 \times(-122)+32] \\
& =[(-219.6)+32] \\
& =-187.6
\end{aligned}
$
So,$-122^{\circ} C$ in Fahrenheit scale is $-187.6^{\circ} F$
Explanation: Substituting the given value for ${ }^{\circ} C =(-122)$ in the following expression,
$
\begin{aligned}
& { }^{\circ} F=\left[1.8\left({ }^{\circ} C\right)+32\right] \\
& =[1.8 \times(-122)+32] \\
& =[(-219.6)+32] \\
& =-187.6
\end{aligned}
$
So,$-122^{\circ} C$ in Fahrenheit scale is $-187.6^{\circ} F$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Which of the following statements $(s)$ is (are) correct
→Which of the following reaction is a redox reaction
The geometry of ammonia molecule can be best described as
Incorrect statement is
Product $(P)$ is
→Which one of the following compounds of Group$-14$ elements is not known?
→Which of the following dissolves in water to give a neutral solution
A weak acid, $HA$ is found to be $10\%$ ionized in $0.01\, M$ aqueous solution. Calculate the $pH$ of a solution which is $0.1\, M$ in $HA$ and $0.05\, M$ in $NaA.$
→A compound ' $X$ ' is a weak acid and it exhibits colour change at $pH$ close to the equivalence point during neutralization of $NaOH$ with $CH _{3} COOH$. Compound ' $X$ ' exists in ionized form in basic medium. The compound '$X$' is.
→The wave function $\psi_{n / m m}$ is a mathematical function whose value depends upon spherical polar coordinates $(r, \theta, \phi)$ of the electron and characterized by the quantum numbers $n, l$ and $m_l$. Here $r$ is distance from nucleus, $\theta$ is colatitude and $\phi$ is azimuth. In the mathematical functions given in the Table, $\mathrm{Z}$ is atomic number and $a_0$ is Bohr radius.
→| $column 1$ | $column 2$ | $column 3$ |
| $(I)$ $1$s orbital | $(i)$ $\psi_{n, l, m_l} \propto\left(\frac{Z}{a_0}\right)^{\frac{3}{2}} e^{-\left(\frac{Z r}{a_0}\right)}$ | $image$ |
| ($II$) $2 \mathrm{~s}$ orbital | $(ii)$ One radial node | $(Q)$ Probability density at nucleus $\propto \frac{1}{a_0^3}$ |
| $(III)$ $2 p_z$ orbital | $(iii)$ $\psi_{n, l m_l} \propto\left(\frac{Z}{a_0}\right)^{\frac{5}{2}} r e^{-\left(\frac{Z r}{2 a_0}\right)} \cos \theta$ | $(R)$ Probability density is maximum at nucleus |
| $(IV)$ $3 \mathrm{~d}_{\mathrm{z}}^2$ orbital | $(iv)$ $x y$-plane is a nodal plane | $(S)$ Energy needed to excite electron from $n=2$ state to $n=4$ state is $\frac{27}{32}$ times the energy needed to excite electron from $n=2$ state to $n=6$ state |
($1$) For the given orbital in Column $1$, the only $CORRECT$ combination for any hydrogen-like species is
$[A] (IV) (iv) (R)$ $[B] (II) (ii) (P)$ $[C] (III) (iii) (P)$ $[D] (I) (ii) (S)$
($2$) For $\mathrm{He}^{+}$ion, the only INCORRECT combination is
$[A] (II) (ii) (Q)$ $[B] (I) (i) (S)$ $[C] (I) (i) (R)$ $[D] (I) (iii) (R)$
($3$) For hydrogen atom, the only $CORRECT$ combination is
$[A] (I) (iv) (R)$ $[B] (I) (i) (P)$ $[C] (II) (i) (Q)$ $[D] (I) (i) (S)$
Give the answer quetion ($1$) ($2$) and ($3$)