Q: This sign shows when a lift is safe to use. Total mass of people must be 480 kg or less Fred and some other people are in the lift and their total mass is 565 kg to the nearest 5 kg. Fred, who has a mass of 88 kg to the nearest kg, gets out of the lift. Is the lift now safe to use? Yes/No You must show your working.

The correct answer is: Yes Explanation 565 kg to the nearest 5 means the total mass M lies in the interval $562.5 \le M < 567.5$. Fred's mass f to the nearest kg means $87.5 \le f < 88.5$. After Fred leaves the lift the remaining mass is $M-f$. Steps: Minimum possible remaining mass: $$562.5 - 88.5 = 474.0\text{ kg}$$ Maximum possible remaining mass (worst case): $$\text{(just less than) }567.5 - 87.5 = 480.0\text{ kg}$$ so $$M-f<480.0\text{ kg}$$ Therefore the remaining mass is always less than $480\text{ kg}$ (hence $\le 480\text{ kg}$), so the lift is now safe to use.

Q: What do we say? cordially invite you or cordinally invite you

The correct phrase is "cordially invite you." Explanation "Cordially" is the adverb form of the adjective cordial and correctly modifies the verb invite. "Cordinally" is not a standard English word and is incorrect. Examples: "We cordially invite you to the opening reception." "You are cordially invited to attend." Alternative, slightly less formal options: "warmly invite you," "would be pleased to invite you," or "we invite you."

Q: 1 atm = 1.01325 × 10^6 কেন হয়?

The correct answer is: $1\ \text{atm}=1.01325\times10^{6}\ \text{dyn/cm}^2\ ( =1.01325\times10^{5}\ \text{Pa})$ ব্যাখ্যা تعريف: ঐতিহ্যগতভাবে 1 atm = চাপ যা 0°C-এ 760 mm উচ্চতার পারদ (Hg) স্তম্ভ ধরে রাখে। আধুনিকভাবে 1 atm ঠিকভাবে ধরা হয় $101325\ \text{Pa}$ হিসেবে। পারদের স্তম্ভ থেকে চাপ বের করা হয় হাইড্রোস্ট্যাটিক সূত্র দিয়ে: $p=\rho g h$ যেখানে $\rho$ হল পারদের ঘনত্ব, $g$ স্থানীয় ত্বরণ (standard gravity), এবং $h$ হল উচ্চতা। ধাপগুলো: পারদের ঘনত্ব ও মানগুলো নিলে: $\rho_{\text{Hg}}=13.5951\ \text{g/cm}^3=13595.1\ \text{kg/m}^3$, $g=9.80665\ \text{m/s}^2$, $h=760\ \text{mm}=0.760\ \text{m}$. চাপ: $$p=\rho g h =13595.1\times9.80665\times0.760\approx101325\ \text{Pa}$$ প্যাস্কালকে dyn/cm^2-এ রূপান্তর: $1\ \text{Pa}=1\ \text{N/m}^2=10\ \text{dyn/cm}^2$। তাই $$101325\ \text{Pa}=101325\times10\ \text{dyn/cm}^2=1.01325\times10^{6}\ \text{dyn/cm}^2.$$ অতএব, ঐতিহ্যগতভাবে 760 mmHg (বা সংজ্ঞা হিসেবে $101325\ \text{Pa}$) থাকার কারণে $1\ \text{atm}=1.01325\times10^{6}\ \text{dyn/cm}^2$।

Q: 1 calorie = ____________ joule.

The correct answer is: $4.184\text{ J}$ Explanation 1 calorie (small calorie, symbol cal) is defined as the amount of energy needed to raise the temperature of 1 g of water by 1 °C; by convention this equals $4.184\text{ J}$. Steps: Definition: $1\text{ cal}=4.184\text{ J}$ (For food Calories) $1\text{ Cal (kcal)}=1000\text{ cal}=1000\times4.184\text{ J}=4184\text{ J}$ Therefore, $1\text{ cal}=4.184\text{ J}$ (and $1\text{ kcal}=4184\text{ J}$ for dietary Calories).

Question

Why you put K_ 273 in the conventional

Accepted Answer

Because absolute zero is at −273.15°C, we add 273.15 (often rounded to 273) to convert Celsius to Kelvin: $T(	ext{K}) = T(°	ext{C}) + 273.15$.

Explanation

The Kelvin scale is an absolute temperature scale whose zero point (0 K) is absolute zero — the theoretical temperature where particles have minimal thermal motion.  
Absolute zero equals −273.15°C, so to move from the Celsius origin (0°C) to the Kelvin origin you must add 273.15 degrees. That gives the exact conversion formula:

$T(	ext{K}) = T(°	ext{C}) + 273.15$.    Conversely: $T(°	ext{C}) = T(	ext{K}) - 273.15$.

Why sometimes people use 273 instead of 273.15

273 is a convenient approximation for quick/rough calculations (error of 0.15 K, usually negligible for many problems).  
For precise science and engineering you should use 273.15 (or more digits if required). By definition of the Celsius/Kelvin scales, the offset 273.15 is exact.

If you meant a different notation (for example $K_{273}$ or something else), tell me and I’ll explain that specific usage.