Computer Awareness Quiz: Memory and Storage

The correct answer is B, register. A register is built into the CPU itself and holds the word the arithmetic and logic unit is operating on at that instant, so it is read and written in a single clock cycle. Nothing in the machine is closer to the processor, and nothing is faster.

Option A, cache memory, is the second fastest and is the right answer only when registers are missing from the options; it sits between the CPU and main memory and is built from static RAM. Option C, main memory, is the DRAM the processor addresses directly, which is fast but an order slower than cache because every cell has to be refreshed. Option D, the hard disk, is secondary storage with moving platters and heads, so its access time is measured in milliseconds against nanoseconds for the levels above it. The order to memorise is register, cache, main memory, secondary storage, magnetic tape.

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.

The correct answer is C, 4.7 GB. A Digital Versatile Disc uses a shorter wavelength laser and finer pits than a compact disc, so one side with one recording layer holds about 4.7 GB; a double-layer disc of the same size holds roughly 8.5 GB.

Option A, 700 MB, is the capacity of an ordinary compact disc, the figure most often mixed up with the DVD in the examination hall. Option B, 1.44 MB, is the capacity of the 3.5-inch floppy disk, a number still asked because it appears in old syllabuses. Option D, 25 GB, is one layer of a Blu-ray disc, which reads with a blue-violet laser of even shorter wavelength. Learn these four capacities together as a set, because a question often lists all of them and asks for one: 1.44 MB, 700 MB, 4.7 GB and 25 GB.

The correct answer is A, the CPU and the main memory. Cache is a small block of static RAM that keeps the instructions and data most recently used by the processor. Because it answers in a fraction of the time DRAM takes, it hides the gap between a fast processor and slower main memory; a request satisfied from it is a cache hit and one that is not is a cache miss.

Option B describes a disk buffer or disk cache, a different arrangement and not what the term cache memory means in this chapter. Option C is wrong because input from a keyboard is handled by a controller and a small buffer, not by the CPU cache. Option D is wrong for the same reason on the output side, where a printer has its own buffer and the operating system uses spooling. Remember the levels as well: L1 lies inside each core, L2 next, and a larger L3 is shared between cores.

The correct answer is B, virtual memory. The operating system reserves an area of the hard disk, called the swap space or the page file, and moves pages a program is not actively using out to it. The program sees one large address space, so software bigger than the installed RAM still runs, though every swap costs time because a disk is far slower than memory.

Option A, caching, does the opposite: it keeps recently used data in a faster memory nearer the processor rather than extending memory onto a slower device. Option C, spooling, queues jobs such as documents waiting to be printed on the disk so the CPU need not wait for a slow peripheral. Option D, defragmentation, rearranges the scattered pieces of files on a disk so they lie in contiguous blocks and read faster; it does not add to memory at all.

The correct answer is B, flash memory chips. An SSD is built from flash memory, itself a block-erasable form of EEPROM, which traps charge in floating-gate cells. Because there is no motor, no platter and no head, an SSD has no seek time, makes no noise and stands up to being knocked about, which is why it has replaced the hard disk in most new machines.

Option A describes the hard disk drive, where data is written magnetically on spinning platters along tracks, sectors and cylinders. Option C describes optical media, the CD, DVD and Blu-ray disc, where a laser burns and reads pits and lands on a reflective layer. Option D describes magnetic tape, the archival medium and the standard example of sequential access storage. All four are secondary storage: the SSD differs in technology, not in its place in the memory hierarchy.

The correct answer is B, 1.44 MB. The 3.5-inch high-density floppy disk, the square plastic diskette with a metal shutter, was formatted to 1.44 MB, and that figure is still asked in computer awareness papers because it anchors the small end of the capacity ladder.

Option A, 720 KB, was the capacity of the earlier double-density 3.5-inch disk, a real figure but not the standard one, and it is the distractor most often chosen. Option C, 2.88 MB, belonged to an extra-high-density format that never came into common use. Option D, 10 MB, was the size of early hard disks, not of any floppy. Keep the set in order and no version of this question can trouble you: 1.44 MB for the floppy, 700 MB for the CD, 4.7 GB for the single-layer DVD and 25 GB for a Blu-ray layer.

The correct answer is B, EEPROM. Electrically Erasable Programmable Read Only Memory is erased by applying a voltage, one byte at a time, while the chip stays soldered in place, which is why it is used for settings that must survive a power cut and yet be changed occasionally, such as the configuration data of a device.

Option A, EPROM, is also erasable and non-volatile but needs ultraviolet light through a quartz window, which erases the whole chip at once and usually means taking it out. Option C, PROM, can be programmed only once by the user and never erased, since programming burns its internal links permanently. Option D, SRAM, is not read only memory at all: it is a fast volatile memory used as cache and loses everything the moment power stops. Note also that flash memory is EEPROM erased in blocks rather than bytes.