Which of the following is the fastest memory in a computer system?

The correct answer is B, 4 bits. A nibble is half a byte, that is four binary digits, and it matters because one nibble is written as exactly one hexadecimal digit, from 0 to F. That is why memory addresses in hexadecimal are read four bits at a time.

Option A, 2 bits, has no standard name in this ladder and is simply a distractor. Option C, 8 bits, is one byte, the space needed to store a single character in the ASCII scheme, and it is the most common wrong choice because candidates confuse the byte with the nibble. Option D, 16 bits, is two bytes, often called a half word on a 32-bit machine. Fix the small ladder in memory: 1 bit, 1 nibble of 4 bits, 1 byte of 8 bits, and then every higher unit multiplying by 1,024.

The correct answer is B, DRAM. Dynamic RAM stores each bit as a charge on a very small capacitor, and that charge leaks away in a few milliseconds, so the memory controller has to read and rewrite every cell continuously. The word dynamic in the name refers to exactly this refresh cycle.

Option A, SRAM, stores a bit in a flip-flop of transistors that holds its state as long as power is supplied, so it needs no refreshing at all; that is what makes it fast enough for cache and too costly for main memory. Option C, ROM, is non-volatile and keeps its contents even without power, so refreshing does not apply. Option D, flash memory, is also non-volatile: it traps charge in a floating gate that holds for years, which is why pen drives and SSDs survive being unplugged. Only dynamic RAM is refreshed.

The correct answer is B, ultraviolet light. An Erasable Programmable Read Only Memory chip carries a small quartz window over the silicon; ultraviolet light shone through that window releases the trapped charge from every cell at once and returns the chip to its blank state, after which it can be programmed again.

Option A, an electrical signal, erases EEPROM, the electrically erasable version, and this is the single most common confusion in the chapter: EPROM needs light, EEPROM needs a voltage, and EEPROM can be erased one byte at a time without removing the chip. Option C, a magnetic field, erases magnetic media such as tape or a floppy disk, not a semiconductor ROM. Option D, infrared radiation, does not carry the energy needed to free the trapped electrons and is only a distractor. Remember also that MROM is written by the maker and PROM can be written once.

The correct answer is C, 1,024 GB. Every step in the ladder of storage units multiplies by 1,024, because 1,024 is 2 raised to the power 10 and memory is addressed in binary. So 1,024 bytes make a kilobyte, 1,024 kilobytes a megabyte, 1,024 megabytes a gigabyte and 1,024 gigabytes a terabyte.

Option A, 100 GB, would be right only if the units advanced in powers of ten, which is how disk makers advertise capacity but not how examiners count it. Option B, 512 GB, is a common drive size and a power of two, but it is half a binary terabyte, not a whole one. Option D, 2,048 GB, is two terabytes. Learn the chain beyond the terabyte as well, because it is asked: petabyte, exabyte, zettabyte and yottabyte, each again 1,024 times the one before it.

The correct answer is B, RAM. Random Access Memory holds a bit only while it is powered, whether the cell is a flip-flop in static RAM or a leaking capacitor in dynamic RAM. Cut the supply and everything in RAM is gone, which is why unsaved work is lost in a power failure.

Option A, ROM, is the standard non-volatile primary memory and keeps the bootstrap loader and the BIOS ready for the next start-up. Option C, the hard disk, stores data as permanent magnetic patterns on its platters, so it needs no power to remember. Option D, the pen drive, uses flash memory, which traps charge in a floating gate and holds it for years without a supply. In short, only RAM among the four is volatile; cache, being static RAM, is volatile too, and is the answer when RAM is not listed.